Oscar Vega

CSET II

Revised July 05, 2011

vi

Copyleft 2011 by Oscar Vega

Copyleft means that unrestricted redistribution and modification are permitted, provided that all copies andderivatives retain the same permissions. Specifically no commercial use of these notes or any revisionsthereof is permitted.

Contents

1 Proofs and Constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Direct proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Contradiction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Basic plane Euclidean geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Postulates and constructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.1 Common Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Non-Euclidean geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

4 Congruency and similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

5 Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.1 General properties of polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Regular polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6 Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.1 Basic properties of triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.2 Triangles and areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.3 Congruency of triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.4 Similarity of triangles and trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

7 Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

8 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478.1 Basic properties of circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478.2 Tangents, secants and chords of circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

vii

viii Contents

9 3-D geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559.1 Planes and lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559.2 Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

10 The Cartesian plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

11 Conic sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

12 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

13 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7113.1 Simple probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7113.2 Probability with multiple events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7213.3 Counting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

14 Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7714.1 Analysis of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7714.2 Hypothesis testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8114.3 Curve fitting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8414.4 Calculator Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

14.4.1 TI-83 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8614.4.2 HP 9g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

Preface

This set of lecture notes cover most of the topics you need to study when you prepare to take the CSET II.Most subjects are looked at in a very deep but straightforward way, which means that the notes might seemdry and too abstract. The idea is that once you understand the concepts covered in these notes then you canspend a good amount of time doing busy work solving examples and practice tests so you can succeed inyour test. Moreover, if you understand concepts instead of just knowing how to solve problems then youwill be able to teach Geometry at a high level and be a teacher who can inspire students by doing things theright way.

It is my believe that just reading these notes is not enough to pass the test, as quite a few topics need tobe discussed and studied with the help of an instructor, or somebody else who knows, and understands, thematerial completely.

Good luck preparing for your test,

O.V.

ix

Chapter 1Proofs and Constructions

One of the important parts of the CSET’s is the constructed response part. These questions weight fourtimes a regular question, and approximate 40% of the whole test. Since, most people agree, one shouldscore above 70% to pass the test then failing the constructed response questions means sure test failing.Moreover, besides the four constructed response questions there are other questions that measure yourability of explaining the reason something is true, and not necessarily to know ‘how to solve’ something,and this is exactly what knowing how to prove something will allow you to do well.

The need for knowing how to write procedures, and express ideas, in proper mathematics makes thisfirst chapter a very important one. In it you will be introduced to proofs, which is the formal name of whatcolloquially one could call ‘a thorough and complete explanation, or deduction, of a fact by using logic’.

Probably the best way to get familiarized with proofs is to read a lot of them and do a lot of them. So,when you read this book do not think proofs as something to skip, but as examples. Doing this will take thefear you might have of proofs out of you, and it will help you to understand better the concepts you need toknow.

As mentioned above, the more you read and do proofs, the better. For the reading part, this book suppliesmany proofs, at different levels of difficulty, and about many different subjects. Most of the times wheneveran important result is mentioned, it is accompanied with its proof. Please read these proofs, understandthem, enjoy them.

For the doing part, there are many exercises at the end of chapters. Most of them are of the form “provethat...”. Please practice your proving skills as, even if you get lucky and not get many proofs in your test, theunderstanding of where things come from will help you to apply those results in other, more computational,problems.

We will look at two proving techniques, a third one will be discussed in chapter 10.

1.1 Direct proof

In a perfect world, all proofs will be obtained as a logic succession of deductions that will lead the argumentfrom the information given (called hypothesis) directly to the desired result (called theorem, lemma, propo-sition, goal, etc). Since we do not live in a perfect world then we will need to learn later about contradiction,but for now let us see how these chains of deductions (AKA direct proofs) can be constructed.

Theorem 1.1 (Vertical Angle Theorem). In the following picture, in which four angles are formed by twointersecting lines

1

2 1 Proofs and Constructions

l

!"

#

m

$

it is always true that α = γ and that β = δ .

Proof. Note that α +β = 180◦ and that γ +β = 180◦. Since both of these sums equal 180◦ then we can setthem equal to each other. We get

α +β = γ +β

We subtract β both sides and get α = γ , which is what we wanted.In a similar way we obtain β = δ . ut

Proofs are not only geometric. For instance, next is a proof about properties of integers.

Theorem 1.2. The sum of two even numbers is also even.

Proof. Consider two even numbers a and b. Since they are even, then they are a multiple of two. In otherwords a = 2n and b = 2m, and n and m are two whole numbers. Hence,

a+b = 2n+2m = 2(n+m)

which means that a+b is a multiple of two, and thus even (that is what we wanted to show). ut

In the next one we will need to have some knowledge about triangles in order to finish the proof. We willlook at two proofs, using the same ideas but expressed in different ways.

Theorem 1.3. An isosceles right triangle must have two 45◦ angles.

Proof. Since the sum of the angles of a triangle is 180◦ then, the triangle already having a 90◦ angle, itmust have two other angles adding up to 90◦. Since an isosceles triangle has (at least) two angles that havethe same measure then the angles that are not right must have the same measure (this is because a trianglecould not have two right angles). It follows that the small angles must measure 45◦. ut

Proof. Let α,β and γ be the three angles of the given triangle. By hypothesis γ = 90◦. Since the sum ofthe angles of a triangle is 180◦ then, α +β = 90◦. Now we use that an isosceles triangle has (at least) twocongruent angles to get that α = β = 45◦. Note that we are using that the triangle could not have two rightangles (see exercise 1.1). ut

1.2 Contradiction

Let us suppose we want to show that a property P is true. When we prove by contradiction we will assumethat P is false, and we will use logic, deductions, etc until we get to something that is impossible (thingslike 1 = 0, even = odd, negative = positive, etc). Since we are reaching something clearly false then ourassumption of P being false cannot be right, this forces P to be true!! Hence, we have reached our goal(proving that P is true) without using direct proof.

Let us look at a couple of examples.

Theorem 1.4. A triangle has at most one obtuse angle.

1.3 Constructions 3

Proof. By contradiction. Let us assume that a triangle 4ABC has more than one obtuse angle. Let α andβ be two obtuse angles of 4ABC. Since α +β > 180◦, then the sum of the angles of 4ABC is more than180◦. This is impossible!

It follows, by contradiction, that at most one angle of4ABC can be obtuse. ut

Another example.

Theorem 1.5. Two lines with a common perpendicular must be parallel.

Proof. By contradiction. Assume that there are two lines, ` and m, that have a common perpendicular t andthat are not parallel. Let C be the point of intersection of ` and m, and let A and B the points of intersectionof ` and m with t, respectively.

Note that4ABC has two right angles (at A and B). This is impossible!! (see exercise 1.1).By contradiction, we get that two lines having a common perpendicular must be parallel. ut

A classical proof by contradiction follows.

Theorem 1.6.√

2 is not a rational number.

Proof. By contradiction. Assume that√

2 is a rational number, and thus

√2 =

ab

where a and b are whole numbers with no common factors (we want the fraction to be in least terms).By squaring and cross multiplying we get 2b2 = a2. Since the number on the left is a multiple of two,

then so must a be. It follows that a = 2n for some whole number n. Let us plug a = 2n into 2b2 = a2. Weget,

2b2 = (2n)2

which isb2 = 2n2

but this forces that b is a multiple of two. Impossible!! a and b cannot be both multiples of two, as they haveno common factors.

By contradiction,√

2 is not a rational number. ut

1.3 Constructions

A construction of a shape, polygon, or figure with certain given properties is the base of geometry, as bydoing constructions one can assure that the objects we are studying and learning about really do exist, andalso discover things that had not been caught before.

Even though constructions may not be considered as proofs, the structure and general idea of not doing,or claiming, something that is not fully justified is present in both proofs and constructions.

In order to construct what is asked to us we need to follow a few simple rules, these are called postulates, and are discussed in chapter 3. For now, without really discussing what the postulates are we will look ata couple of simple constructions to illustrate how to construct shapes using a straight edge and a compass(which means we will construct shapes by drawing lines and circles only).

Example 1.1. Let us construct an equilateral triangle with a given base.Given segment AB with length a. First draw circles with centers A and B and radius a. The intersection

of these circles is called C (note we have two options for C, choose either). Draw lines to create 4ABC,which must be equilateral because of the radius considered for the circles.

4 1 Proofs and Constructions

C

A B

Example 1.2. Assume you know how to construct a perpendicular line to a given line at a given point. Wewant to construct a square given one side of it.

We will just give the construction in words, as an exercise, you should take a straight edge and a compassand perform the construction with the instructions given.

Given segment AB with length a. At A and B construct perpendicular lines to AB. We will use only theparts of these lines that are ‘above’ AB. At A and B draw a circle with radius a. These circles will intersecttheir corresponding perpendiculars in exactly one point. Label these points C (‘above’ A) and D (‘above’B), and join A with C, B with D, and C with D with lines. The shape ABCD is a square with side a.

Example 1.3. We want to construct a 60◦ angle with vertex at a given point A. As in the previous example,you should perform the construction following the instructions given next.

Draw any line through A and a circle with any radius (call it a) centered at A. The line and the circlewill intersect in two points, choose one and label it B. So far we have constructed a segment with length a.By example 1.1 we can construct an equilateral triangle with base AB. Since equilateral triangles have 60◦

angles at each vertex, then we have just constructed the angle at A that we wanted to get.

Many more constructions will be shown in this book. Keep in mind that, just as in the last two examples,previously known constructions can be used to do new, more complex, constructions. Practice this skill, itis a very important one to have.

Problems

1.1. Prove that a triangle has at most one right angle.

1.2. What is the set of points that are all at the same distance from a fixed point C?

1.3. Construct a triangle with sides 2 units, 4 units and 5 units.Hint: Use the longer segment as the base and then use circles with radii 2 and 4 to find the third vertex.

1.4. Can you construct a triangle with sides 2 units, 4 units and 7 units?

Chapter 2Basic plane Euclidean geometry

Before getting into anything complex, in terms of geometry, we need to set what will be the objects we willuse to do geometry. These concepts are probably known by you, thus their descriptions will be short and,sometimes, appealing to your intuition or previous knowledge.

We will think of a point as a dot on a piece of paper. A point has no length or width, it just specifies anexact location. It is interesting that even though a point is almost nothing all geometric shapes are collectionsof points.

We may think of a line as a ‘straight’ line that we might draw with a ruler on a piece of paper, exceptthat in geometry, a line extends forever in both directions. A line passing through two different points A andB is written as

←→AB. Note that the idea of ‘line’ and ’straight’ are linked, and that if one uses one to define the

other (as we did above) then probably we would use the other to define the one. This is not quite correct,but we will (ab)use our intuition in this definition.

Three or more points are said to be collinear if there is a line that contains them. In the picture, the linel contains the points P,Q and R. Hence, P,Q and R are collinear

lP Q R

Since, there is always a line through any two given points, then we could say that two points are alwayscollinear.

Two lines either intersect or they are parallel. Note that this could be used as a definition of parallel lines:lines that do not intersect. Three or more lines are concurrent if they all pass through the same point. In thefollowing picture, l,m and n are concurrent.

n

l m

A ray is the portion of a line that has one endpoint and extends indefinitely from the endpoint on. A raywith endpoint A and passing through a point B is written as

−→AB or

←−BA.

The rays−→AB and

−→AC are opposite rays if they are distinct and the points A,B and C are collinear. Opposite

rays form a 180◦ angle.A line segment is the portion of a line that is between two points (the two points included). A line

segment with endpoints A and B is written as AB. The length of the segment AB is the distance betweenA and B. Two segments, AB and CD, having the same length are said to be congruent. We denote that asAB∼=CD.

Remark 2.1. Given two distinct points, the distance between them is always positive (in particular neverzero).

5

6 2 Basic plane Euclidean geometry

A point P between A and B such that AP∼= PB is called the midpoint of the segment AB. The midpointis said to bisect the segment.

Remark 2.2. The existence of a midpoint will be given later by finding a way to explicitly construct it fromA and B. Moreover, given a segment, its midpoint is unique.

Proof (Uniqueness of a midpoint). Assume there are more than one (distinct) midpoint of AB, call two ofthem P and Q. We know, by remark 2.1 that the distance between P and Q is not zero, but this is impossible,as both are equidistant from A and B. Contradiction. ut

An angle with vertex A is a point together with two rays−→AB and

−→AC (called the sides of the angle)

emanating from A. We call this angle ∠BAC or ∠CAB. Often we will use lowercase greek letters to denoteangles.

It is very customary to identify an angle with its measure. Be careful with this, as the measure of anangle only captures part of what the angle really is. In fact, two angles with the same measure are said to becongruent.

We will mostly use the sexagesimal system to represent angles. That is, we will consider angles tohave a measure between 0◦ and 360◦. However, there are other ways to measure angles. Radians are themost used in trigonometry, calculus, etc. In this system and angle of 180◦ corresponds to π radians. Thiscorrespondence defines a proportionality between these twi types of measures.

For example, if you want to know how many radians is 45◦, then by using proportions one gets

180◦

π=

45◦

x

Solving for x one obtains that a 45◦-angle also measures π

4 radians.

Remark 2.3. Given two distinct rays with a common endpoint, the angle formed by them has always posi-tive measure.

A bisector of a segment is a line that passes through the midpoint of the segment. If the bisector intersectthe given line forming a 90◦ angle, then it is called a perpendicular bisector.

In the picture l is the perpendicular bisector of PQ, and R is the midpoint of PQ

R

l

P Q

Remark 2.4. There is a very simple way to construct the midpoint and/or the perpendicular bisector of agiven segment AB by using just an unmarked ruler and a compass.

One first draws two circles with the same radius centered at A and B, The radius must be large enoughso the circles intersect at two points. The line joining these two points is the perpendicular bisector of AB,and thus passes through the midpoint of AB. We will see later why this works, we first need to learn a fewthings about triangles.

Definition 2.1. Two angles are said to be congruent if one of them could be placed on top of the other fora perfect match. Congruent angles have the same measure.

Two angles with a common vertex and that share a side are said to be adjacent angles. Two nonadjacentangles formed by two intersecting lines are called vertical angles.

2 Basic plane Euclidean geometry 7

Remark 2.5. Theorem 1.1 says that two vertical angles must be congruent (Vertical Angle Theorem orVAT).

An acute angle is an angle whose measure is greater than 0◦ and less than 90◦. A right angle hasmeasure exactly 90◦. An obtuse angle is an angle whose measure is greater than 90◦ and less than 180◦. Astraight angle has measure exactly 180◦.

Two angles are called complementary if the sum of their measures is exactly 90◦. Two angles arecalled supplementary if the sum of their measures is exactly 180◦. Note that two complementary (orsupplementary) angles need not to be adjacent. A linear pair of angles are adjacent angles whose non-common sides are opposite rays (form a straight line), these angles must be supplementary.

An angle bisector is a ray whose endpoint is the vertex of the angle and which divides the angle intotwo congruent angles.

In the picture,−→BD is a bisector of ∠ABC if and only if ∠ABD∼= ∠DBC.

D

B

A

C

Remark 2.6. There is a nice and simple way to construct the angle bisector of a given angle.This construction uses the one we did for the perpendicular bisector of a segment, and thus the details

will be clarified later when we go over triangles.First you draw a circle centered at the vertex of the angle, this circle intersect the sides of the angle in

one point each. Call these points A and B. It turns out that the perpendicular bisector of AB goes throughthe vertex of the angle and, in fact, is the angle bisector we were looking for.

Two lines ` and m are perpendicular if they intersect at a point P and if there is a ray that is part of ` anda ray that is a part of m that form a right angle. Perpendicular lines ` and m are denoted by `⊥ m.

Remark 2.7. The distance between a point P and a line ` is the length of the segment that starts at P andhits ` in a right angle. That is, the distance from P to ` is given by a perpendicular line to ` through P.

Two lines in the same plane which never intersect are called parallel lines. We say that two line segmentsare parallel if the lines that they lie on are parallel. If line `1 is parallel to line `2 we write `1||`2.

Remark 2.8. If one throws a perpendicular to, let us say, `1 at ANY point P ∈ `1, then the segment created‘between’ the lines `1 and `2 has always the same length, independently of the point P... in other words,two parallel lines are always equidistant.

Let l1 and l2 be two lines and m a transversal to both of them forming eight angles, like in the followingpicture,

m

! "

#

$ %

& '

(

l1

l2

We can see that we get pairs if angles that look congruent (for example, α and ε). If any of these pairs isa pair of congruent angles then l1||l2, and α = ε = δ = ϕ and β = γ = φ = ψ .

Conversely, if l1||l2 then α = ε = δ = ϕ and β = γ = φ = ψ .

8 2 Basic plane Euclidean geometry

Problems

2.1. In the proof of remark 2.2. Explain with a picture why the distance between P and Q being not zerocontradicts that both points are equidistant from A and B.

Note that you are using that the midpoint(s), A and B are collinear. Explain why this should be true.

2.2. Prove that given a segment, its perpendicular bisector is unique.Hint: Use remark 2.3 and the idea used to prove remark 2.2.

2.3. Prove that, given an angle, its angle bisector is unique.

2.4. Given 4 points in a plane. How many lines are determined? Note that considering different locationsfor the points will yield different answers.

2.5. The intersection of two rays might be.....What about their union?

2.6. Suppose K,L, and M are on a line `, with L between K and M. Which term best describes the set of allpoints P such that L is between M and P ? (assume that L could be equal to P).

2.7. How many lines are determined by the ten points in the diagram? (Points that appear to be collinear arecollinear)

2.8. Show that there are exactly 12 positive angles less than 360◦ in the figure.

2.9. Find the measures in radians of the angles 0◦, 30◦, 60◦, 90◦, 270◦, and 360◦.

2.10. Find a point P such that AP∼= PC. Then find a point Q such that AB∼= BQ

C

0 2 4 6 8!2

A B

2.11. An angle’s measure is three times the measure of its supplement. What is the measure of the angle ?What is its complement?

2.12. Consider the following picture.

2 Basic plane Euclidean geometry 9

B

AC

D l

!

"

where l is the bisector of the angle ∠ABC. Prove that α = β .

Chapter 3Postulates and constructions

Euclid wrote his famous book, The Elements, about 23 centuries ago. His idea was to summarize all themathematical knowledge of his times by obtaining results (theorems) from known ones, and this knownones were to be obtained also from other, more basic, known results, etc. Proceeding in this fashion, heeventually got to as set of very obvious, or self-evident, statements that would be the base of everything(math of course)!! He called these evident statements Common Notions and Postulates. Of course, beforewhat follows, Euclid listed all necessary definitions. We will not define everything in detail, as most of theconcepts discussed here are known to us.

What follows is, essentially, taken from book I of Euclid (that explains the weird way it is all written.Although I have changed the phrasing so it is all easier to understand). Visit

htt p : //aleph0.clarku.edu/∼d joyce/ java/elements/elements.html

for a more complete review (with pictures and Java applets. Very nice stuff.)

3.1 Common Notions

These are more numerical or algebraic, but are necessary for what follows.

1. Things which equal the same thing also equal one another.2. If equals are added to equals, then the wholes are equal.3. If equals are subtracted from equals, then the remainders are equal.4. Things which coincide with one another equal one another.5. The whole is greater than the part.

Note that we use these notions every day when we solve equations and inequalities. However, the phras-ing is so vague that, for example, ‘the whole’ is not necessarily a number or algebraic expression, it couldalso be an angle, the area of a shape, the length of a segment, etc.

3.2 Postulates

Now is when the geometry starts. These postulates assume that all elements considered (points, lines, etc)are on the same plane and that all lines have infinite length.

11

12 3 Postulates and constructions

1. It is possible to draw a straight line from any point to any point. (Every two points determine a uniqueline).

2. It is possible to extend a segment continuously into a whole line.3. It is possible to describe a circle with any (known) center and (known) radius.4. All right angles equal one another. (A right angle had been defined by Euclid as an angle that is

congruent to its supplement).5. If a straight line falling on two straight lines makes the interior angles on the same side less than two

right angles, the two straight lines, if produced indefinitely, meet on that side on which are the anglesless than the two right angles.The ‘translation’ of this postulate says that since in the picture

!

"

n

m

l

the sum of the angles α and β is less than 180◦, then l and m, once extended into whole lines, willintersect ‘on the side’ of n where α and β are.

Since Euclid wrote The Elements there were people thinking that the fifth postulate was special, it wasn’tvery ‘self-evident’, and clearly was much more complicated than the previous ones. In fact, it is believedthat Euclid himself though the fifth was odd, as he held the use of it as long as possible; the fifth postulateis used for the first time in proposition 29.

For centuries the greatest mathematical minds attempted to determine whether the fifth was really apostulate, all of them were unsuccessful until in the nineteenth century Nikolai Lobachevsky and JanosBolyai (separately) proved that the fifth postulate was independent from the previous 4 postulates (it is also‘known’ that Gauss had already figured this out himself, but chose not to publish his work). They did thisby constructing geometries where the first 4 postulates hold but not the fifth, their examples (essentially thesame) were the first examples of non-Euclidean geometries.

But, what is a non-Euclidean geometry? By definition it is a geometry (points and lines on a plane, andall one can construct out of them) where the fifth postulate is false, but the other four are true. We will seelater a few properties of these new geometries.

Since Euclid did not use the 5th postulate until proposition 29 then the first 28 propositions are valid notonly in the ‘standard’ Euclidean geometry but also in all non-Euclidean geometry, it is customary to callNeutral Geometry to the geometry (together with all its results) that assumes the first four postulates butdoes not assume anything about the fifth.

Now we will go over the arguments that prove a few of these propositions that do not need the fifth.These proofs/constructions are very important for you, as they are good examples of questions you mightget in the constructed response part of the CSET.

1. How to construct an equilateral triangle given its base (constructed in example 1.1).2. How to place a segment congruent to a given segment with one end at a given point. i.e. How to copy a

segment

Proof. Given segment AB and a point P. First join P and A with a line (postulate 1). Using proposition1 we can find Q such that4APQ is equilateral. Now draw a circle centered at A with B on the boundary(postulate 3).

3.2 Postulates 13

E

A B

P

Q

D

Extend the segments QA andn QP into lines (postulate 2), call D and E to the intersections of these linesand the circle already constructed. Now draw a circle centered at Q with D on the boundary (postulate3).Note that AB∼= AD and that QD∼= QE. Since QA∼= QP, it follows that AB∼= AD∼= PE.

E

A B

P

Q

D

ut

3. How to cut off from the larger of two given unequal segments a segment congruent to the shorter.

Proof. Using the previous proposition we copy the shorter segment and put its beginning at one extremeof the longer segment. It follows that we can now cut off (literally) the shorter segment from the longer.

ut

4. SAS criterion for congruence of triangles (proved as theorem 6.3).5. The base angles of an isosceles triangle are congruent (proved as theorem 6.1).6. If in a triangle two angles are congruent to each other, then the sides opposite the equal angles are also

congruent to each other (proved as theorem 6.1).Note that this is the converse of the previous proposition.

8. SSS criterion for congruence of triangles (proved as theorem 6.4).9. How to bisect a given angle α .

Proof. We first draw a circle centered at the vertex V of α .

14 3 Postulates and constructions

R

V!

P

Q

The intersections of this circle with the sides of the angle are labeled P and Q. Now we draw two circleswith the same radius centered at P and Q. These circles intersect at R (note we have two choices for R,choose either).Note that the triangles 4V PR and 4V QR are congruent by SSS. It follows that ∠PV R ∼= ∠QV R. Thus−→V R is the bisector of α . ut

10. How to bisect a given segment.

Proof. The segment AB is given. We draw circles with the same radius centered at A and B. Thesecircles intersect in two points, which we label C and D.

C

M

D

A B

The line through C and D intersects AB at a point M. We claim that M is the midpoint of AB. Moreover,that the line

←→CD is the perpendicular bisector of AB.

The claim follows from the fact that the points C and D are equidistant from A and B and thus4CBD∼=4CAD. It follows that ∠BCD∼=4ACD, which implies that4CMB∼=4CMA by SAS. Hence AM∼=MB.The fact that

←→CD is the perpendicular bisector of AB follows from the fact that the angles ∠AMC is

congruent to ∠CMB and that their sum is 180◦. ut

11. How to draw a line at right angles to a given line from a given point on it.12. How to draw a line perpendicular to a given line from a given point not on it.15. If two straight lines cut one another, then they make the vertical angles equal to one another.

This is the vertical angle theorem. It was proved as theorem 1.1.16. In any triangle, if one of the sides is extended, then the exterior angle is greater than either of the interior

and opposite angles.

Proof. We want to show that in the picture below ∠CBD is larger than both ∠CAB and ∠ACB.C

A B D

3.3 Non-Euclidean geometry 15

Using proposition 10 we find the midpoint of CB, call it E. Then we draw the line from A to E, and usingproposition 2, and postulate 2, we find a point F on it such that AE ∼= EF . We obtain the picture

C

A

G

E

F

DB

where E is the midpoint of both CB and AF . It follows that4AEC ∼=4FEB by SAS, and thus ∠ACB∼=∠FBE, which is smaller than ∠EBD.Similarly, we obtain that ∠CAB is smaller than ∠EBD. ut

Note that in the proof above, we need to find the point F by extending a line until it reaches a desiredlength. This is done by assuming that lines extend indefinitely. Hence, for this proposition to work weneed infinite lines.

26. ASA criterion for congruence of triangles (proved as theorem 6.5).Note that if we know that two triangles have two pairs of corresponding congruent angles, then the thirdangle of one triangle must be congruent to the third angle of the other. Hence, ASA can be also phrasedas AAS or SAA.

27. If a line is transversal to two lines making the alternate angles congruent to one another, then the straightlines are parallel to one another.

28. If a line is transversal to two lines making the exterior angle congruent to the interior and opposite angleon the same side, or the sum of the interior angles on the same side equal to 180◦, then the lines areparallel to one another.

The previous two propositions say that, in the following picture, if γ = ψ or γ = φ , then l1 is parellelto l2

m

! "

#

$ %

& '

(

l1

l2

As mentioned before, all the results above do not require the fifth postulate. But there are many, many,MANY others that do. In fact, most of what we will see in this course depends in one way or another onthis postulate.

3.3 Non-Euclidean geometry

In order to see what we obtain when we decide to assume that the fifth postulate is false, we need to look atother better-known results that turn out being equivalent to the fifth. Here there are a few

1. (Playfair’s axiom) Let ` be a line and P a point not on `, then there is exactly one line passing through Pthat is parallel to `.

16 3 Postulates and constructions

2. The sum of the interior angles of a triangle is exactly 180◦.3. (Proposition 29) If in figure used for propositions 27 and 28 we assume l1||l2, then the alternate interior

angles are equal, and the corresponding angles are congruent.4. The area of a triangle is half its base times its height.

Now that we are more familiar with what the fifth means we can see what it means to assume it is false.We will focus mostly on Playfair’s axiom, as the negation of it gives us two options

(i) Let ` be a line and P a point not on `, then there are more than one line passing through P that are parallelto `.In this case we obtain what is called hyperbolic geometry. An example of this is the hyperbolic plane (itsdescription is too complex this course).

(ii) Let ` be a line and P a point not on `, then there are no lines passing through P that are parallel to `In this case we obtain what is called elliptic geometry. An (almost) example of this is the geometry ofthe sphere, where the lines are the great circles (meridians, equators, etc).

The following table shows how things work in the three geometries.

Euclidean Hyperbolic Elliptic

The Fifth True False False

Proposition 29 True False False

Number of parallel lines to ` through P /∈ ` 1 ∞ 0

Angle-sum of a triangle 180◦ < 180◦ > 180◦

It depends It depends

Area of a triangleb ·h

2on the on the

angle-sum angle-sum

Problems

3.1. Prove propositions 11 and 12.Hint: Use the idea used in the proof of remark 2.2 after obtaining an interval on the given line using the

point given.

3.2. Assume the angle sum of any triangle is 180◦

(a) Show proposition 27.(b) Show proposition 28.

3.3. In this exercise you will prove that the angle sum of any triangle is 180◦

(a) Consider a triangle4ABC and use Playfair’s axiom to get a line through C that is parallel to←→AB.

(b) Extend the three sides of the triangle to obtain the picture

3.3 Non-Euclidean geometry 17

C

! "

#

AB

l

l

1

2

Use proposition 29 to determine the angles at C, and then get the desired result.

3.4. Choose two points on a sphere/ball. Draw the shortest path on the sphere that joins these points. Con-vince yourself that this ‘line’ must be a great circle of the sphere.

3.5. Draw a triangle on a sphere/ball. Note that on the sphere lines are ‘curvy’, and that triangles formedwith these lines are like a swollen (Euclidean) triangle. Use a protractor to measure the angle-sum of thistriangle. It should be more than 180◦, is it?

Repeat this exercise with many different triangles on the sphere. Do you note any relation between theangle-sum of the triangles and their area?

3.6. Show that (in Euclidean geometry) the sum of two angles of a triangle is equal to the opposite exteriorangle of the triangle. Show this is not true in non-Euclidean geometry.

3.7. A man comes out of his house, walks 10 miles South, then 10 miles East, and finally 10 miles North.He is now back at his place. When he is getting inside he sees a bear. What color is the bear?

3.8. If the sum of the interior angles of a triangle is equal to 200◦ then how many parallel lines to a line `can be found through a point P not on `?

What if the sum of the interior angles of a triangle were 100◦?

Chapter 4Congruency and similarity

When we have two figures/shapes that overlap perfectly once one of them is placed on top of the other wewill say that the shapes are congruent. The symbol for ‘congruent to’ is ∼=. The idea behind this concept isthat there are movements of the plane that preserve shapes, angles, distances, etc. These movements will bediscussed later when we learn about transformations in chapter 12.

In case there is a movement of the plane that preserves shapes and angles but not distances then thesemovements are essentially ‘zooming’ (in or out) one shape into the other. Two figures that are related thisway are said to be similar. The symbol for ‘similar to’ is ∼.

We can make these concepts more precise for polygons and other familiar shapes.

Definition 4.1. Two given polygons are said to be congruent if by creating a one-to-one correspondencebetween their vertices then we get that all pairs of corresponding angles and all pairs of corresponding sidesare congruent.

Similarly, in order to get a definition of similarity of polygons we need the fact that when we zoom in/out(to check similarity) we use the same scaling for all distances on the plane, the scale used can be obtainedby dividing the lengths of corresponding sides. That is, by the ratio of two corresponding sides.

Definition 4.2. Two polygons are said to be similar if there is a one-to-one correspondence between theirvertices such that all pairs of corresponding angles are congruent and the ratios of the measures of all pairsof corresponding sides are equal.

Remark 4.1. If two shapes are congruent then they are similar. Two similar shapes need not be congruent.

In the following proposition we summarize a few easy results about congruency and similarity,

Proposition 4.1. 1. Two segments are congruent if and only if they have the same length.2. Any two segments are similar. The ratio is given by dividing the length of one by the length of the other.3. Two angles are congruent if they have the same measure.4. It two angles are similar then they are congruent (similarity preserves angles).5. Two circles are congruent if they have the same radius.6. Any two circles are similar. The ratio is given by dividing the radius of one by the radius of the other.

The correspondence of vertices mentioned in definitions 4.1 and 4.2 is very important, as with it we showin which way we are looking at the polygons that are congruent/similar. Mislabeling vertices could yield towrong results. For instance, in the following picture4ABC ∼4PQR but4ABC 6∼ 4RPQ.

PA

CQ

B R

19

20 4 Congruency and similarity

One important thing to remark about congruency and similarity is that they are transitive. That is, if A,Band C are figures/shapes such that A∼= B and B∼=C, then A∼=C. Similarly, if A∼ B and B∼C, then A∼C.

One way to think about similarity is that two shapes are similar if one can be zoomed into the other. Buthow does this work? We are used to say things as ‘zoom ×2’ but what does that mean? What is it doubled?Lengths !! For example, a square of side 3 in when zoomed ×2 becomes a square of side 6 in. Note that thearea of the square does NOT double but actually quadruples (4 = 22). Similarly, a cube of side 3 in whenzoomed ×2 becomes a cube of side 6 in. Note that the volume of the cube is multiplied times 8 = 23.

Since there is not much to do so in general (for ALL shapes, or ALL polygons), then we will move onto other sections, where we will discuss congruency or similarity for specific shapes.

Problems

4.1. I want to build a pool in my yard. It must have the same shape as the yard itself but (of course) smaller.In fact, I want the perimeter of the pool and the perimeter of the yard to be in a ratio of 1 : 2. What wouldbe the ratio of the areas enclosed by the pool and the yard?

4.2. Prove proposition 4.1.

4.3. True or False? If true, prove it. If false, give an example that supports your claim.(a) Same area squares are similar.(b) Same area squares are congruent.(c) Same area rectangles are similar.(d) Same area rectangles are congruent.(e) Same area circles are similar.(f) Same area circles are congruent.

Chapter 5Polygons

Polygons are the most basic closed shapes in geometry. They can be large/complex enough to be interestingbut since they have only segments (no curves) by sides then they are fairly easy to study. Probably the mostimportant polygon is the triangle, as it will be present in most of the techniques we will discuss to learnabout polygons. Due to their importance, triangles have their own chapter (chapter 6).

5.1 General properties of polygons

Definition 5.1. A polygon is a set/shape formed by points (called vertices) and segments (called sides) suchthat every vertex is the endpoint of exactly two sides.

Note that the previous definition implies that a polygon is a closed shape. Also, under that definition ashape like

is considered to be a polygon (in this case with 5 vertices and 5 sides). However, this is not what comes tomind when we think of a polygon, as the star above has sides intersecting in points that are not vertices, wedo not want this kind of behavior in the polygons we want to study.

Definition 5.2. A polygon with sides intersecting only at vertices is called a simple polygon.

Remark 5.1. In this course we will always assume polygons are simple.

Now, let us consider the following simple polygon

A

D B

C

which, then again, might not be what we usually consider as a polygon. What we do not line about the shapeabove is that it is not convex (that interior angle ∠DCB is too large!!).

21

22 5 Polygons

Definition 5.3. A simple polygon is said to be convex if the segment joining any two of its points is com-pletely contained in the ‘frame’ given by the polygon.

Another way to think about a convex polygon is as a simple polygon that has interior angles measuringat most 180◦.

From now, in this book, all polygons are convex simple polygons.Probably the first way to characterize a polygon is by counting how many sides it has. Although this is

not enough to say much about the polygon, it is something that helps the visualization of the shape we wantto study.

Definition 5.4. (i) When we say ‘side AB’ we will mean the side that joins vertices A and B. Note that thisimplies that once all the vertices of the polygon are labeled, then there is a unique way to label all its sided.(ii) A polygon having n sides is sometimes called an n-gon.(iii) The perimeter of a polygon is the sum of the lengths of all the sides of the polygon.

The second feature that is used to characterize a polygon is its angles, they come in two flavors: theinterior angles of a polygon are the angles (inside the polygon) formed by two consecutive sides of thepolygon. The exterior angles of a polygon are the angles obtained by extending one side (either) at allvertices.

!

b a

f

ed

c

"

#

$%

&

In the picture above, a,b,c,d,e, and f are internal angles, and α,β ,γ,δ ,ε and φ are external angles.

Proposition 5.1. The sum of the interior angles of a polygon with n sides equals n ·180◦ minus the sum ofthe exterior angles.

Proof. Since the exterior angles are obtained by extending a side, then an exterior angle is always thesupplement of an interior angle (and adjacent to that angle as well). Since there are n interior angles and foreach one of them we get a linear pair of angles formed by one interior angle and one exterior angle, thenthe sum of the measures of all angles (interior and exterior) must be n ·180◦.

Separating interior and exterior angles we get

(sum of interior angles) + (sum of exterior angles) = n ·180◦

Subtracting yields what the proposition claims. ut

A diagonal is obtained by joining two distinct (and not consecutive) vertices in a polygon. We can seethat a triangle has no diagonals, and that a quadrilateral has exactly two. The following result generalizesthese numbers to any (simple convex) polygon.

Proposition 5.2. The number of diagonals in an n-gon isn(n−3)

2.

Proof. We count the number of diagonals by fixing a vertex (we have n to choose from), then we multiplythis by the number of options for the second extreme of the diagonal (there are n−3 available). Thus we getn(n−3) diagonals. But since the same two extremes determine the same diagonal, then we need to divideby two to get the correct number. ut

Since polygons can have any number of sides, and each side be of, pretty much, any length, then it isimpossible to get a through and general study of all polygons. Hence, we will focus on polygons that aremore ‘predictable’, with have more symmetry in them.

5.2 Regular polygons 23

5.2 Regular polygons

Every triangle that has three angles having the same measure must have three sides with the same length.This is not true in polygons with more than three sides. For instance, a rectangle (that is not a square) hasfour 90◦ angles but it does not have four sides with the same length. Similarly, a rhombus (see chapter 7)has four sides with the same length but its angles do not have all the same measure.

Definition 5.5. An equilateral polygon is a polygon with all sides congruent to each other.An equiangular polygon is a polygon with all interior angles congruent to each other.If a polygon is equilateral and equiangular then the polygon is regular.

Example 5.1. As mentioned above. A rectangle that is not a square is equiangular but not equilateral (andthus not regular). A rhombus that is not a square is equilateral but not equiangular (and thus not regular). Atriangle that is equiangular must be regular, and a triangle that is equilateral must be regular.

A regular polygon possesses a point C (called the center of the polygon) that is equidistant from allvertices of the polygon. The distance r is called the radius of the polygon. The height of the trianglesformed by using the center is called the apothem of the polygon (denoted ρ or just a).

a C

r

In fact, the triangles formed by throwing all the radii are isosceles and congruent to each other.Using the idea shown in the picture above, we see that a regular n-gon can be broken into n congruent

isosceles triangles. If we add the angles in these triangles we get n ·180◦, but since by counting all the anglesin the triangles we also count the angles that are formed around the center of the polygon (which add up to360◦), then we get that the sum of the interior angles of the polygon is

n ·180◦−360◦ = (n−2)180◦

This idea may be generalized to any polygon, and this would imply that the sum of the exterior angles is360◦.

But if the polygon is regular we can go further, now we can learn that the measure of each interior angleis

(n−2)180◦

nMoreover, the measure of each exterior angle of a regular n-gon is

180◦− (n−2)180◦

n=

360◦

n

which is not so surprising, as the sum of the n exterior angles of the polygon is 360◦.

24 5 Polygons

ar

Remark 5.2. In the picture above it is shown that the apothem is the radius of the circle inscribed in thepolygon, this circle is tangent to each side of the polygon at its midpoint. Also, the radius of the polygon isthe radius of the circle circumscribed to the polygon, this circle touches the polygon only at its vertices.

Note that the more sides a regular polygon has, the closest it is to be a circle, and thus a and r to be thesame number. Archimedes used this idea to find good approximations for the value of π , among many otherthings.

Recall that the area of a triangle isb ·h

2, where b is the base of the triangle and h its height. In the case

of a regular n-gon, with side s and apothem a, we see that the area of the polygon is n times the area of oneof the isosceles triangles formed by the radii. It follows that the area of the n-gon is

A = n · s ·a2

= a · s ·n2

=a ·P

2

where P is the perimeter of the polygon.The previous formula is great to compute the area of a regular polygon if you know its perimeter (or just

the length of one of its sides), but the problem is that we do not know any way to find the apothem of aregular polygon (unless it is given to us, which does not happen often).

Proposition 5.3. If α is an interior angle of a regular polygon and s is the length of a side of the polygon,then the apothem a is given by

a =s2

tan(

α

2

)The function ‘tan’ will be introduced in chapter 6, in the trigonometry section. So far, you can use your

scientific calculator to compute values for tan if needed.

Note that now we can find all the important features of a regular polygon, as long as we know its numberof sides and the length of a side. Moreover, if one considers two n-gons with sides s and t, then these twopolygons are similar, in a ratio s : t, and their apothems and radii are in the same ratio.

Problems

5.1. Show that the sum of the interior angles of an n-gon equals n ·180◦ minus the sum of the exterior anglesof the polygon.

5.2. Assume that we know a polygon can be partitioned into triangles that share one vertex, look at thepictures below. Use this to find what the sum of the interior angles of the polygon is.

5.2 Regular polygons 25

5.3. If in a regular polygon you double the length of the sides, then what happens with the perimeter andarea of the polygon?

5.4. What is the perimeter of the figure below?

28

24

(All angles are right)

5.5. What is the degree measure of an interior angle in a regular octagon? An exterior angle?

5.6. In the picture below, both hexagons are regular and have the same center. Also, the big one has apothemtwice the small one’s. Find the area of the shaded region.

r

a

r

a

5.7. The measure of an interior angle in a regular polygon is 120◦. Find the number of sides of the polygon.

5.8. Show that the six triangles formed by the radii and the sides of a regular hexagon are equilateral.

5.9. How many sides does a regular polygon have if the measure of an interior angle is three times themeasure of an exterior angle?

Chapter 6Triangles

6.1 Basic properties of triangles

1. A triangle is a polygon with exactly three sides. Depending on the picture one can determine what thebase of the triangle is. For instance, the triangle4ABC below

A B

C

has base AB. Also, the angles ∠CAB and ∠ABC are called base angles.2. A triangle with at least two sides congruent is called an isosceles triangle. As a convention, we will

represent an isosceles triangle as

A B

C

where AC ∼= BC.

Theorem 6.1. The base angles of an isosceles triangle are congruent. And, conversely, a triangle havingcongruent base angles is isosceles (these are propositions 5 and 6).Also, the altitude of an isosceles triangle bisects the base in a right angle, and it is also the angle bisectorof the angle opposite to the base.

Proof. Call P the point of intersection of the angle bisector of ∠ACB and AB. This creates two triangles,4APC and 4BPC. Since Since AC ∼= BC, ∠ACP ∼= ∠BCP, and PC is common to both triangles, then4APC ∼=4BPC (by SAS, which we will prove later). It follows that the base angles are congruent, that∠APC ∼= ∠BPC, and thus that PC is perpendicular to the base of the triangle.Moreover, if4ABC above has congruent base angles (but we don’t know yet it is isosceles), then callingQ to the point where the angle bisector of ∠ACB meets the base. Then we get two triangles 4AQC and4BQC, which are congruent by ASA (which we will prove later)... as ∠ACQ∼=∠BCQ, ∠CAQ∼=∠CBQ,and thus ∠CQA∼=∠CQB and the segment QC is common to both triangles. It follows that AC∼= BC, andthus the triangle would be isosceles. ut

27

28 6 Triangles

Remark 6.1. The previous proof can be modified to show that the points on the perpendicular bisectorof a segment AB are equidistant from A and B. In fact the perpendicular bisector can be re-defined usingthis property, thus the perpendicular bisector of AB is the set of all points equidistant from A and B.

3. All interior angles of an equilateral triangle are congruent. That is, and equilateral triangle is a regular3-gon. This follows from the fact that an equilateral triangle is also isosceles.

4. A triangle with a right angle is called a right triangle. The two sides that form the right angle are calledarms or legs of the triangle, the third side is called the hypothenuse.

5. A triangle with one obtuse angle is called an obtuse triangle.6. A triangle with three acute angles is called an acute triangle.7. A median of a triangle is a segment from a vertex to a midpoint of its opposite side.8. An altitude of a triangle is a segment starting at a vertex and ending at the closest point on the line

containing the opposite side. Altitudes are always perpendicular to the line containing the opposite side.9. The segment that joins the midpoints of two sides of a triangle is called a midline of the triangle.

Remark 6.2. A midline’s length is one half of the length of the third side of the triangle. Also, a midlineis parallel to the third side. That is, in the picture

C

A F B

E

the length of AC is twice the length of EF , and AC||EF .We will soon see that4ABC ∼4FBE.

10. The perpendicular bisectors of a triangle are the three perpendicular bisectors of the sides of the trian-gle.

Remark 6.3. The perpendicular bisectors are concurrent, the intersection point (called the circumcen-ter) is the center of the circle circumscribed to the triangle (see pictures below)

Why is this true? Note that two perpendicular bisectors always intersect, take this intersection point andcall it C. Using remark 6.1. we get that this point is equidistant to the three vertices of the triangle, whichmeans that C is the center of a circle going through the three vertices of the triangle.Finally, to see that the third perpendicular bisector also goes through C we just need to see that the linethat goes through C and the midpoint of the side we haven’t used yet contains (at least) two points thatare equidistant to the vertices of the side, thus using the paragraph above we get that this line (through Cand the midpoint) is the third perpendicular bisector we were missing.

11. The angle bisectors of a triangle are the three bisectors of the angles of the triangle.

Remark 6.4. The angle bisectors are concurrent, the intersection point (called the incenter) is the centerof the circle inscribed to the triangle (see picture below)

6.2 Triangles and areas 29

The proof to the previous remark is similar to the proof of the existence of the circumcenter. You shouldbe able to do it.

6.2 Triangles and areas

1. The Pythagorean theorem: In a right triangle with legs measuring a and b and hypothenuse measuringc, then a2 +b2 = c2

There are hundreds of proofs for this theorem, read this for example

htt p : //math f orum.org/library/drmath/view/62539.html

A couple of them are exercises in problem list 2. You should know them well. Also, read the file aboutthis that is posted on the course’s website (The 2500-year old Pythagorean theorem).

2. The area of a triangle with base b and altitude h is A(4) =bh2

.The proof for this follows from the fact that once a triangle is given we can ‘double’ it to create aparallelogram

h

bSince the area of the rectangle is bh then the area of the triangle is half of that.Note that ‘doubling’ the triangle to create a parallelogram means that the two triangles forming theparallelogram are congruent. Also, we will show later that the area of that parallelogram is actually basetimes height.

3. Another way to find the area of a triangle is what is called Heron’s formula:

Theorem 6.2. The area of a triangle with sides with length a,b, and c is given by

A(4) =√

s(s−a)(s−b)(s− c) where s =a+b+ c

2

Proof. We first rotate the triangle (if necessary) so the point where the altitude intersects the base parti-tions it into two segments, p and q, as shown in the picture below.

cA B

C

ab

p q

h

30 6 Triangles

We use the Pythagorean theorem in the two triangles created by the altitude to get

h2 + p2 = b2 and h2 +q2 = a2

Since p+q = c then

q2 = (c− p)2

= c2−2cp+ p2

a2−h2 = c2−2cp+ p2

which implies (using h2 + p2 = b2) that

a2 = c2−2cp+b2 and thus p =b2 + c2−a2

2c

Using that we get,

c2h2 = c2(b2− p2)

= c2(b− p)(b+ p)

= c2(

b− b2 + c2−a2

2c

)(b+

b2 + c2−a2

2c

)= c2

(2cb− (b2 + c2−a2)

2c

)(2cb+(b2 + c2−a2)

2c

)=

14(a2− (b2−2cb+ c2)

)((b2 +2cb+ c2)−a2)

=14(a2− (b− c)2)((b2 + c)2−a2)

=14(a+b− c)(a−b+ c)(b+ c+a)(b+ c−a)

We now use that

a+b− c = 2(s− c), a−b+ c = 2(s−b), b+ c+a = 2s, b+ c−a = 2(s−a)

and that the square of area of the triangle is A =h2c2

4to get the formula we wanted. ut

6.3 Congruency of triangles

Recall that (chapter 4) two triangles are congruent if there is a correspondence between the vertices of twotriangles that yields congruent correspondent sides and congruent corresponding angles. So, in theory, inorder to check that two triangles are congruent we need to check six congruences (three for sides, and threefor angles), that is too much to check. Luckily, there are congruence criteria for triangles that only ask forthree things to check.

Theorem 6.3 (SAS). If two triangles have two sides congruent to two sides respectively, and have the anglecontained by the congruent sides congruent, then the triangles are congruent.

Proof. The proof for this is by constructing a triangle with the given information, and realizing that therecould be only one such a triangle.

Assume that two lengths, a and b, and the angle α between them are given, thus we have a picture like.

6.3 Congruency of triangles 31

b

!

a

It is pretty clear that there is a unique way to complete that picture to create a triangle. Done.

Theorem 6.4 (SSS). If a triangles has its three sides congruent to the sides of a second triangle, then thetriangles are congruent.

Proof. This proof is similar to the previous one. We assume three lengths are given: a, b, and c. We set c asthe base of the triangle (with extremes A and B), and then we draw circles centered at A and B with radii aand b. We get the following picture.

a

cABb

We note that the two points of intersection of the circles are both at distance b from A and a from B.Thus these two points are the only candidates to be the third vertex of the triangle we want to get.

Now I will just say that these two triangles are congruent (finishing the proof), as one of them is thereflection of the other

c

b

b a

a

BA

... however, it is not that easy to prove that.

Theorem 6.5 (ASA). If two triangles have two angles congruent to two angles respectively, and one sideequal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equalangles, then the triangles are congruent. Note that is a little more than just ASA.

Proof. First of all we realize that since knowing two angles in a triangle immediately tells us what the thirdangle MUST be, then it is enough to show ASA.

So, we assume we know two angles α and β and their common segment a. This yields the followingpicture.

! "a

which clearly show that there is a unique way to obtain a triangle with that information.

32 6 Triangles

Now a couple of classical constructions that use the latest three results. We sort of discussed these beforebut without so much detail.

I How to bisect a given angle α .

Proof. We first draw a circle centered at the vertex V of α . The intersections of this circle with the sidesof the angle are labeled P and Q. Now we draw two circles with the same radius centered at P andQ. These circles intersect at R (note we have two choices for R, choose either). Note that the triangles4V PR and4V QR are congruent by SSS. It follows that ∠PV R∼= ∠QV R. Thus

−→V R is the bisector of α

R

V!

P

Q

II How to bisect a given segment.

Proof. The segment AB is given. We draw circles with the same radius centered at A and B. Thesecircles intersect in two points, which we label C and D. The line through C and D intersects AB at a pointM. We claim that M is the midpoint of AB. Moreover, that the line

←→CD is the perpendicular bisector of

AB.

C

M

D

A B

The claim follows from the fact that the points C and D are equidistant from A and B and thus4CBD∼=4CAD. It follows that ∠BCD∼=4ACD, which implies that4CMB∼=4CMA by SAS. Hence AM∼=MB.The fact that

←→CD is the perpendicular bisector of AB follows from the fact that the angles ∠AMC is

congruent to ∠CMB and that their sum is 180◦.

6.4 Similarity of triangles and trigonometry

1. The definition of similarity tells us that if4ABC ∼4A′B′C′ then

ABA′B′

=AC

A′C′=

BCB′C′

where AB means the length of AB (similar for the others)... and that all corresponding angles are congru-ent, which means that

∠ABC ∼= ∠A′B′C′ ∠BCA∼= ∠B′C′A′ ∠CAB∼= ∠C′A′B′

6.4 Similarity of triangles and trigonometry 33

Recall that 4ABC ∼ 4A′B′C′ and 4CAB ∼ 4A′B′C′ might mean different things because the corre-spondence of vertices/sides/angles is determined by the way things are written.Since triangles are fairly simple, we can find criteria for similarity, just as we did with congruency oftriangles. We have two main results:

i If two triangles have two pairs of corresponding angles that are congruent, then they are similar (alsocalled AA).

ii If two corresponding sides are in the same ratio and the angles they form (in each triangle) are con-gruent, then the triangles are similar (this is some sort of SAS with ratios).

Remark 6.5. The triangle formed by the midlines of a triangle is similar to the original triangle.

This is true because once parallel lines are thrown, and the sides of the smaller triangle are extended weget

!

!

!

!

which creates a correspondence between vertices of the big and small triangles. The same can be donefor all the other angles. One gets,

!"

"

#

#

!

So, the triangles are similar by AA.

Remark 6.6. The ‘triangle inside another triangle’ and the ‘bowtie‘ figure are classical cases of similartriangles. In fact, if

B

A

C

D

E

where AB||DE, then4ABC ∼4DEC. And if

34 6 Triangles

A

B

C

D

E

where AC||DE, then4ABC ∼4EBD.

2. Now we will look at similarity of right triangles. Note that since two right triangles have both a rightangle, then as soon as they have a second angle ‘in common’ (meaning congruent) then they will besimilar. Consider the following two similar right triangles,

2

! !A

OH

A1

11

O 2

2

H

where the A’s indicate ‘adjacent to α’, the O’s denote ‘opposite to α’, and the H’s denote ‘hypothenuse’.By similarity we get

A1

H1=

A2

H2

O1

H1=

O2

H2

O1

A1=

O2

A2

Since these ratios depend only on the angle α , then they are functions of α , we give them the followingnames

sinα =OppHyp

cosα =Ad jHyp

tanα =OppAd j

Let us find the sine and cosine of some important angles. For instance, using the isosceles triangle

1

1

we can find the third side to be√

2 by using the Pythagorean theorem, then we can compute

sin45o =1√2=

√2

2and cos45o =

1√2=

√2

2

Now consider a right triangle4ABC that is equilateral and with CD to be one of its altitudes/perpendicularbisectors/angle bisectors like in the picture below

30

A

B

C D

2

2

1

1

60

6.4 Similarity of triangles and trigonometry 35

Using the Pythagorean theorem we get that the altitude’s length is√

3. Now looking at the triangle4CDB we can find that

sin30o =12

cos30o =

√3

2

sin60o =

√3

2cos60o =

12

A handy little table to recall some important values of the main two trigonometrical functions

0o 30o 45o 60o 90o

sin√

0 1 2 3 42

cos√

4 3 2 1 02

3. Now consider the (not necessarily right) triangle4ABC

!

A B

C

ab

c" #

Theorem 6.6 (Law of sines). In the previous picture the following holds

asinα

=b

sinβ=

csinγ

Proof. We first provea

sinα=

bsinβ

by dropping the altitude from C in the previous picture.

A B

C

ab

! "

h

Then,

sinα =hb

sinβ =ha

So, solving for h in these equations and setting them equal to each other we get

bsinα = asinβ

which implies what we want.The other part of the formula is proved similarly.

Theorem 6.7 (Law of cosines). In4ABC (see picture above) the following holds

c2 = a2 +b2−2abcosγ

36 6 Triangles

This law yields two more ‘formulas’ by rotating the triangle, they are

a2 = b2 + c2−2bccosα and b2 = a2 + c2−2accosβ

Proof. Clearly, it is enough to show

b2 = a2 + c2−2accosβ

We look at the picture we used in the proof of the law of sines, and we notice that the base of the trianglehas been cut into two pieces. We can find the lengths of these pieces by using the cosine of β , we get.

A B

C

ab

! "

h

a cosc ! a cos" "

Using the Pythagorean theorem we get

a2 = h2 +(acosβ )2 b2 = h2 +(c−acosβ )2

Solving for h2 in both equations and setting them equal to each other we get

a2− (acosβ )2 = b2− (c−acosβ )2

which impliesa2− (acosβ )2 = b2− c2 +2accosβ − (acosβ )2

Canceling (acosβ )2 we geta2 = b2− c2 +2accosβ

which implies the result we wanted.

4. (Thales’ theorem) Let4ABC be the right triangle in the picture below

DA Bq

h

p

C

where h is the height, p is the length of AD and q is the length of DB.Then, since4ADC ∼4CDB by AA, taking ratios of corresponding sides yields

ph=

hq

or h2 = pq

Another cool property of right triangles is that, referring to the picture above, if M is the midpoint of ABthen the segment CM is exactly half as long as AB, thus after drawing this segment we have 4ABC asthe union of two isosceles triangles.

Problems

6.1. Let 4ABC be an isosceles triangle with base AB and AC ∼= BC. Prove that the triangle with base ABand sides given by the two angle bisectors of the base angles of4ABC is isosceles.

6.4 Similarity of triangles and trigonometry 37

6.2. Oscar wants to find the height of a tree. He is exactly six feet tall, and notices that at the time he castsan eight-foot shadow. He measures the tree’s shadow and finds that it is 74 feet long. How tall is the tree?

6.3. Agnes wants to tie a support line from the top of a 50 foot radio tower to an anchor spot 30 feet fromthe towers base. Approximately how long will the line need to be? (You might need a calculator for thisone)

6.4. The triangle4ABE is inscribed within square ABCD and has a height of 6 cm. Assuming that the littletriangle inside4ABE is formed by midlines. What is the area of the shaded region?

ED

AB

C

6.5. According to the following figure

!"

#

D

E

C

Assuming that α = 65◦. Give all pairs of congruent angles. Then give all pairs of similar triangles.

6.6. Assume that AC bisects ∠ECB, EB bisects ∠ABC, and that AB is parallel to EC

A

C

B

D

E

Show that4CDB is right.

6.7. Consider the following right triangle

b

ac

Now take four copies of it and arrange them in a square in the following way

38 6 Triangles

c

b a

a

b

a b

a

b

cc

c

First show that the square-like polygon that is contained in the large square is actually a square. Thencompute the area of the big square in two ways. First by using the length of its side and then by computingthe areas of the polygons (small square and triangles) that form it. Conclude the Pythagorean theorem.

Now with the same right triangle used above, use the following figure to find a different proof for thePythagorean theorem.

b

c

a

cc

c

a

bb

b

a

a

6.8. Does Angle−Side−Side work as a criterion for congruence of triangles?

6.9. Is there a triangle with sides 2 in, 4 in, and 7 in?

6.10. We know that if a right triangle has sides measuring a, b, and c then a2 + b2 = c2. Show that if atriangle has sides measuring a, b, and c, and a2 +b2 = c2 then the triangle must be right.

6.11. Assume that AD is the angle bisector of ∠CAB and that BD is the angle bisector of ∠ABC.

D

A B

C

Show that twice the measure of ∠ADB equals the measure of ∠ACB plus 180◦.

6.12. Assume that AB is parallel to DE

E

A B

C

D

Show that if ∠CDE and ∠ABC are complementary then4ABC is right.

6.4 Similarity of triangles and trigonometry 39

6.13. Let ABCD be a rectangle with diagonals AC and BD. Show that4ABC ∼=4DCB.

6.14. Show that any triangle 4ABC can be broken into four congruent triangles, each one of them similarto4ABC in a 1 : 2 ratio.

6.15. In the figure below. Assume that EB⊥ AC and AD⊥CE and AB∼= DE

EA

B

C

D

F

Show that4CAD∼=4CEB.

6.16. I suspect the front and back of my tent are not congruent. So, I look at it against the sum and see

F

A B C D

E

where any two segments that seem to be parallel are, in fact, parallel, and that AB is congruent to CD.Am I right assuming the back and front of my tent are not congruent?

Chapter 7Quadrilaterals

1. A polygon with 4 vertices (and sides) is called a quadrilateral.2. Two sides of a quadrilateral that have a common vertex are called adjacent sides and two sides that do

not have a common vertex are called opposite sides.3. A parallelogram is a quadrilateral whose opposite sides are parallel.4. Two angles that have their vertices at the endpoints of the same side of a parallelogram are called con-

secutive angles.5. An altitude of a parallelogram is a segment that is perpendicular to two opposite sides (or the lines that

contain them).6. The area of a parallelogram with altitude measuring h and base b is A = bh.

It is easy to see that a parallelogram with width larger than height has the area described above. Ifthe parallelogram is a ‘skinny’ parallelogram, then the argument is a little more complex but it worksanyway.

7. A rectangle is a parallelogram with all interior angles measuring 90◦.

Remark 7.1. The diagonals of a rectangle are congruent. The converse is also true, i.e. if a parallelogramhas congruent diagonals, then it must be a rectangle.

Remark 7.2. It is in the definition that the interior angles of a rectangle are right, but it is also true thatif a quadrilateral has four right angles then it must be a rectangle.

8. A rhombus is a quadrilateral with four congruent sides.

Remark 7.3. The diagonals of a rhombus are perpendicular to each other. Moreover, they are anglebisectors.The converses of these two facts are also true but only when the quadrilateral is already a parallelogram.In fact, for a parallelogram to be a rhombus it is enough that one diagonal is a bisector.Now look at the figure below (a kite), in it we see that the diagonal DB is a bisector and that the twodiagonals intersect in 90◦, however ABCD is not a rhombus, as their sides are not congruent.

D

A

B

C

41

42 7 Quadrilaterals

9. The area of a kite is given by

A =d1d2

2where d1 and d2 are the lengths of the diagonals.

Proof. Consider the kite in the figure above, where d1 is the length of the horizontal diagonal, and d2 thelength of the vertical one.Since AB∼= BC, AD∼= DC, and BD is common to4BAD and4BCD, then4BAD∼=4BCD. It followsthat the area of the kite is twice the area of4BAD, which is

d2 · d12

2=

d1d2

4

The result follows. ut

10. A square is a quadrilateral that is both a rhombus and a rectangle.

Remark 7.4. If a rhombus has one right angle, then it is a square.

Remark 7.5. Since a square is a rhombus, then we can compute its area using its diagonals.

11. A trapezoid is a quadrilateral with only two sides parallel. The two parallel sides of a trapezoid arecalled bases (sometimes I might call one the base and the other the summit).

12. The two angles adjacent to the base are called base angles, the two angles adjacent to the summit arecalled summit angles.

Remark 7.6. Two consecutive angles in a trapezoid, not both base or summit angles, are suplementary.

This is because the side of the trapezoid defines a transversal to the two bases, which are parallel.13. An altitude of a trapezoid is a segment that is perpendicular to both bases (or to the lines containing

them).14. The area of a trapezoid is given by

A =h(b1 +b2)

2where h is the length of the altitude, and b1 and b2 represent the two bases.

Proof. Consider the trapezoid

We extend the base (summit) the length of the summit (base), then we join the end points to create thefigure

Since the base and the summit are parallel, then it is easy to show that the two trapezoids in the picture(one upside down) are congruent and form a parallelogram. The are of this parallelogram is (b1 +b2) ·h,where b1 and b2 are the length of the base and summit of the trapezoid, and h is the height of thetrapezoid. Since the parallelogram has twice the area of the trapezoid we are done. ut

15. The median of a trapezoid is the segment joining the midpoints of the two non-parallel sides.

7 Quadrilaterals 43

Theorem 7.1. The median is parallel to the bases and it measures exactly

b1 +b2

2

Proof. Consider A trapezoid ABCD with median LM. Draw two perpendicular lines to AB (and thus,also to CD) from L and M. Finally, extend CD to a line. We obtain the following figure.

SA B

CDP

L M

Q

R

Since AL∼= LD (L is the midpoint of AD), ∠ALR∼= ∠PLD (vertical angles), and both ∠DPL and ∠LRAare right, then4ALR∼=4DLP by SAS. Similarly,4CQM ∼=4BSM.It follows that PL ∼= LR and QM ∼= MS, and thus LM is a median of rectangle RSQP. Hence, LM isparallel to the summit and base of the rectangle (which are the same lines giving us the base and summitof the trapezoid).We also get that LM ∼= PQ∼= RS, and thus LM is one-half of the sum of the lengths of PQ and RS. But,since 4ALR ∼=4DLP and 4CQM ∼=4BSM then the sum of the lengths of PQ and RS is equal to thesum of the lengths of DC and AB ut

Remark 7.7. (i) If m is the length of the median of a trapezoid with height h, then we can say that thearea of a trapezoid is A = mh.(ii) Another proof of theorem 7.1 may be obtained by using the techniques in chapter 10.

16. An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent.

Remark 7.8. In an isosceles trapezoid the two base angles are congruent, and so are the two summitangles.

This follows from the picture below

where the two red lines are heights. Since the trapezoid is isosceles, then the two triangles are congruent(they are right triangles with two congruent corresponding sides). So, the base angles of the trapezoidare congruent.

Remark 7.9. The diagonals of an isosceles trapezoid are congruent.

If you draw the diagonals you will obtain two congruent triangles, formed by the base of the trapezoid,a diagonal and a side. Hence, the diagonals are congruent.

Remark 7.10. Any two opposite angles in an isosceles trapezoid are suplementary.

This follows from the fact that the two summit angles are congruent and the two base angles are congru-ent, and remark 7.6.

17. In case you are like me and forget so many properties about trapezoids, just keep in mind the followingpicture.

44 7 Quadrilaterals

which shows how a trapezoid can be ‘extended’ to a triangle, and since the bases of the trapezoid areparallel, then the big triangle is similar to the small (dashed) one... pretty much all you need to know canbe obtained by from this picture.

Problems

7.1. Oscar wants to cover a football field with sod. The field is 360 feet long and 200 feet wide. Sod canbe purchased in squares in 1 foot increments from 2 feet wide up to 9 feet wide. What is the largest sizesquares Oscar can purchase with which he can cover the field completely without any gaps or overhang?

7.2. What is the perimeter of a rhombus with diagonals measuring 32 in and 24 in.

7.3. Find the height of an isosceles trapezoid with base 22 in, summit 8 in and other two sides measuring25 in.

7.4. The trapezoid RWT S is isosceles

P

R

S T

W

Is4RPW isosceles?

7.5. Let LM be the median of the trapezoid ABCD, and P the midpoint of AD

M

A

B C

DP

L

If LP∼= MP. Show that ABCD is isosceles.

7.6. If α > β . Can ABCD be a rectangle?

7 Quadrilaterals 45

!A B

CD

E

"

7.7. Let #ABCD be a parallelogram, AE ∼=CF ,

!

A

B C

D

E

F

(a) Find ∠CDF in terms of α .(b) Show ∠ABE ∼= ∠CDF .

7.8. Consider the following parallelogram ACDF

E

A B C

DF

(a) Assume that AB∼= ED. Show that BCEF is a parallelogram.(b) Assume that FB and EC bisect ∠AFD and ∠ACD respectively. Show that BCEF is a parallelogram.

Chapter 8Circles

8.1 Basic properties of circles

1. A circle is the set of all points that are at the same distance r from a given point C. The number r is calledthe radius of the circle and C is the center. The largest distance between two points on a circle is d = 2r,which is called the diameter of the circle.

Remark 8.1. Two distinct circles intersect in at most two points.

Proof. We have examples of two circles intersecting in zero, one or two points. Assume two circlesintersect in at least two points. We get the picture

O P

Q

R

where the dashed line (joining the centers of the circles) is the altitude for both isosceles triangles4ROQand 4RPQ. It follows that once the point R (for example) has been determined, then there is a uniqueother point of intersection... given by the dashed line and R. It follows that two circles can intersect in atmost two points. ut

2. A line intersects a circle C in at most two points. A line that does not intersect C is said to be exterior, aline that intersects C in exactly one point is a tangent, and a line intersecting C is called a secant.

3. The segment, created by a secant line, that is between two points of the circle is called a chord.

Remark 8.2. If a chord is perpendicular to a radius, then the radius bisects the chord

Proof. Joining the center with the extremes of the chord we will create a triangle that has altitude theradius that is perpendicular to the chord. Since this triangle is also isosceles, it follows that the altitudemust bisect the base (which is the chord). ut

Remark 8.3. If two chords are at the same distance from the center then they are congruent.

Proof. Recall that distances from a point to a segment/line are measured using segments that are per-pendicular to the original line. In this case the radius is that perpendicular segment. Using the previous

47

48 8 Circles

problem we get that the distance to the center is actually the height of the isosceles triangle created byjoining the center with the extremes of a chord. Since the sides of these triangles are uniquely determined(they are radii) and the height is also unique, then their bases must be uniquely determined. Done. ut

4. Let us recall that there is a second way to measure angles, by using radians. We use radians when wewant to use length (inches, yards, centimeters, etc) to measure angles. The conversion rate from degreesto radians is given by 180◦ = π radians.

5. An angle with vertex at the center of the circle is called a central angle. An angle with vertex at a point ofthe circle is called an inscribed angle. In the picture below ∠ACB is central, and both ∠AEB and ∠ADBare inscribed.

E

!

"

#C

A

B

D

Remark 8.4. Note that the angles in the picture above share points A and B, when this happens we getthat the central angle is exactly twice the inscribed angle. That is,

2γ = 2β = α

Proof. Just consider one inscribed angle and a central angle. Draw a radius from the center to the vertexof the inscribed angle, and AB creating three isosceles triangles.

A

B

D

C

From the picture we get (and using4DAC,4ABC and4BDC are isosceles)

2∠ADC+∠DCA = 180◦

2∠BDC+∠DCB = 180◦

2∠CAB+∠ACB = 180◦

which implies2∠ADC+2∠BDC+∠DCA+∠DCB = 360◦

Also, the three angles around the center add up to 360◦, so

∠DCB+∠DCA = 360◦−∠ACB

Plugging this into the previous equation we get

2∠ADC+2∠BDC+(360◦−∠ACB) = 360◦

8.1 Basic properties of circles 49

which implies2(∠ADC+∠BDC) = ∠ACB

Done. ut

Remark 8.5. Using the previous remark we can see that any triangle inscribed in a circle with the diam-eter as a base must be a right angle. That is, in the picture below, DB is the diameter and both triangles4DBC and4DBA are right.

D

A

B

C

Remark 8.6. Any pair of opposite angles in a quadrilateral inscribed in a circle are supplementary. Thatis, in the picture below ∠D+∠B = ∠C+∠A = 180◦.

D

A

B

C

6. The points on a circle that lie between the two points a central angle intersects the circle is called acircle arc. The whole ‘slice’ obtained by drawing a central angle is called a circle section. So, in the nextpicture, the shaded area is the slice given by the central angle and the blue (continuous line) is the arcdetermined by the central angle.

7. The perimeter (or circumference) of a circle with radius r is P = 2πr. The area of a circle is A = πr2.

Remark 8.7. The length of a circular arc with central angle α (in radians) is `α = αr. The area of a

circle sector with central angles α is Aα =αr2

2.

It follows that congruent central angles yield congruent arcs and sectors.

This is just a simple application of proportions.

50 8 Circles

8.2 Tangents, secants and chords of circles

1. In the picture below, α equals one-half of the sum of the central angles that are determined by the arcsAB and CD.

D

!

A

BC

Proof. Let P be the intersection of BD and AC, and join A with D creating 4APD. Note that ∠PADmeasures half the length of arc CD, and that ∠PDA measures half the length of arc AB. Since α isexterior to4APD with opposite interior angles ∠PAD and ∠PDA. The results follows. ut

2. In the picture below, α equals one-half of the difference of the central angles that are determined by thearcs AB and CD.

!

A

B

C

D

Proof. Let P be the vertex of the angle α . Join A with C creating4APC. Note that ∠ACB is external to4APC. It follows that α +∠CAP = ∠ACB, or

α = ∠ACB−∠CAP

The result follows from the fact that arc AB is held by ∠ACB, and arc CD is held by ∠CAP. ut

3. In the picture below,BE ·ED = AE ·EC

E

A

B

C

D

(note that I am talking about the multiplication of the lengths of those segments)

Proof. Since lengths of segments are in the result, then we suspect we need to get a couple of similartriangles going on here.Join A with B and C with D creating ∠ABD and ∠ACD. Since these angles hold the same arc (AD), thenthey are congruent. Moreover, ∠BEA ∼= ∠CED, thus 4BEA ∼ 4CED. Looking at the ratios of theircorresponding sides we get

8.2 Tangents, secants and chords of circles 51

BECE

=AEED

which implies what we want. ut

4. A tangent line to a circle at a point P is perpendicular to the radius that contains P. That is, in the picturebelow, where ` is tangent to the circle with center C at the point P,

r

C

P

l

the line ` is perpendicular to CP

Proof. If this were not true, then we draw a perpendicular to l falling from C. Call Q to the point ofintersection of these two perpendicular lines. It follows that the (shortest) distance from C to l is thelength of CQ. Thus, the length of CP (which is r) must be larger than the length of CQ. It follows that Qmust be inside the circle, which will force the line to be a secant. This is not possible. Hence CP mustbe perpendicular to l. ut

5. Let us throw two tangent lines from a point A outside the circle

Q

P

A

Then AP∼= AQ.

Proof. Let O is the center of the circle. Note that4OPA and4OQA are right triangles with a commonhypothenuse. Moreover, OP∼= OQ because both are radii of the circle. It follows from the Pythagoreantheorem that AP∼= AQ. ut

6. In the following picture, where←→PR is tangent to the circle at P,

R

P

Q

the angle ∠QPR is one-half the central angle with arc PQ.

Proof. Let O be the center of the circle. If P, O and Q are collinear then the result is clear.Assume P, and O and Q are not collinear. Draw radii OP and OQ, creating an isosceles triangle4POQ.The altitude of this triangle must intersect AP at a point different from P because P, and O and Q are notcollinear. In fact the altitude must bisect PQ. We get this picture.

52 8 Circles

RP

QO

!

"

It follows that ∠PT R is right and thus ∠T PS = α . We are done because ∠POQ = 2α . ut

7. In the following picture

R

P

A

Q

AQAP

=APAR

Proof. Just as like in a previous problem, we need two similar triangles. Looking at the final formula wecan guess what triangles we need to work with. We will show4QPA∼4PRA.We draw segments PR and PQ creating a couple of triangles. Using the previous problem we get that∠PQR∼= ∠RPA.Since ∠PAQ is common to both triangles, then by AA we get4QPA∼4PRA. Done. ut

Problems

8.1. Assume that ABC, ACE, and CDE are semicircles, and that AC∼=CE. Show that the area of the shadedregion is equal to the area of4ACE

D

A E

CB

8.2. Consider the following figure

8.2 Tangents, secants and chords of circles 53

which is obtained by drawing a square inscribed in a circle and then four semicircles with diameters thesides of the square.

Show that the red area is equal to the blue area.

8.3. Show that if two chords of the same circle are congruent then the arcs they determine must be congruentas well.

8.4. Assume that CXEY is a square and that C is the center of the circle. Show that the arc QP is congruentto the arc JT .

J

PT

E

XY

Q C

8.5. Find x

102

3

x

8.6. From a point 2 units from a circle, a tangent is drawn. If the radius of the circle is 8 units, find thelength of the tangent segment.

8.7. Find the circumference of a circle that is circumscribed about a square whose perimeter is 36 in.

8.8. Let p and q be two positive numbers. Prove that

√pq≤ p+q

2

Hint: Use Thales’ theorem (chapter 6) and remark 8.5.

Chapter 93-D geometry

9.1 Planes and lines

1. Let us called E3 to our three-dimensional space. E3 is formed by points, lines, planes and many othercurves and solids.

2. Two distinct lines in E3 can be skew (no intersection and pointing in different directions), parallel (nointersection and pointing in the same direction), or they intersect in a point.

3. Just as an infinite number of points create a line, an infinite number of lines create a plane, which onlyhas two dimensions (width and length).

Remark 9.1. If two distinct planes in E3 intersect, then they do so along a line. Also, two distinct planescould be parallel (meaning no intersection).

4. The normal vector (or line) of a plane is a vector (line) that is perpendicular to all lines on the plane. Forinstance, in the picture below, the normal vector would be the red one (the middle one) and not the othertwo (blue ones)... they just don’t seem to be perpendicular enough!

5. The dihedral angle between two planes is the angle formed by their normal vectors.

9.2 Solids

1. A prism is a solid with parallel congruent bases (most of the times polygons)

Fig. 9.1 Prisms

55

56 9 3-D geometry

A prism with bases that are not aligned one directly above the other is called an oblique prism.

Fig. 9.2 Oblique prism

Remark 9.2. A prism with a polygonal base has parallelograms as lateral faces.

Examples of prisms are: cube, rectangular solid (shoe box), cylinder, etc.2. The surface area of a solid is the sum of the areas of all its sides. The volume of a solid measures the

amount of space it occupies in space.There are many formulas to find the surface area and volume of the most common solids, here it is a list.

Prism Cone Pyramid Sphere

Volume h ·Abase13 hπr2 1

3 h ·Abase43 πR3

Sur f ace area h ·Pbase +2Abase πrs+πr2 ? see below 4πR2

where

a. Abase is the area of the base of the solid,b. Pbase is the perimeter of the base of the solid,c. h is the height of the solid,d. r is the radius of the circle that is the base of the cone,e. R is the radius of the spheref. s is the length of the slant height of a cone (see picture below)

s

? The surface area of the pyramid depends on the number of sides, and the shape, of the polygon usedfor the base.

9.2 Solids 57

Problems

9.1. Two opposite vertices in a rectangular box are at distance 7√

2 in. If the sides of the box are in the ratio1 : 2 : 3, then what is the volume of the box?

9.2. Consider a ‘sand-clock-like’ figure formed by two (not necessarily) congruent cones sharing their ver-tex. This figure is perfectly fit inside a cylinder that has the same base as the cones’... something like thefollowing picture.

What portion of the volume of cylinder is the volume of the sand clock?

9.3. A food company wants to paint their entire tomato paste can red, including the top and bottom. If thecan is a cylinder with lid’s diameter equal to 4 cm, and height 7 cm, what is the total surface area that willneed to be painted?

9.4. Consider a cone that has been cut/sliced by a plane that is parallel to its base, like in the picture below

If the height of the cone is 12, the height of the little cone that is formed above the highest plane is 4,and the slant height of the big cone is 15, then what is the ratio of the surface areas of the cones?

9.5. If the edge of a cube is y in. What is the distance from one corner of the cube to the furthest corner onthe opposite side of the cube?

9.6. The radius of a sphere is tripled, by what number is its volume multiplied?

9.7. If a cube and a 3×8×9 inches rectangular box have the same volume. What is the area of a side of thecube?

9.8. The volume of a cylinder is 5076 f t3. Its base radius and height are in a ratio of 3 : 10. Find the surfacearea of the cylinder.

9.9. A bowl in the shape of a half-sphere with radius 5 in is full of water. If the radius of the base of acylindrical container is 2 in, how tall does it need to be to contain all the water in the bowl?

Chapter 10The Cartesian plane

In this section we will learn enough to use coordinates to prove geometric properties of triangles, quadri-laterals, and many other shapes. We will learn more techniques in coordinate geometry in the section onconics.

1. We can write every point P on the plane as an ordered pair of numbers P = (x,y). The numbers x and yare called the coordinates of P.The converse is also true, given a pair (x,y) then it is possible to graph the pair as a point on the plane.This presentation of the plane as a set of ordered pairs is called the Cartesian plane.

2. Given two points P = (x1,y1) and Q = (x2,y2) then we can find the distance between P and Q by usingthe formula

d(P,Q) =√

(x2− x1)2 +(y2− y1)2

3. The midpoint of the segment PQ, where P = (x1,y1) and Q = (x2,y2) is the point

M =

(x2 + x1

2,

y2 + y1

2

)

4. The equation of a line is given by y = mx+b, where m is the slope of the line and b gives the y-interceptof the line.The slope of the line through the points P = (x1,y1) and Q = (x2,y2) is

m =y2− y1

x2− x1

Remark 10.1. The slope determines the angle the line makes with the x-axis. In fact m = tanα , whereα is the angle formed by a line with slope m and the x-axis.

Remark 10.2. Two lines that are parallel have the same slope.The product of the slopes of two lines that are perpendicular is equal to −1.

With the tools learned above we can show things such as: (1) the midline of a trapezoid is parallel tothe base and its length is equal to one-half of the sum of the bases, (2) if the diagonals of a rectangle areperpendicular, then the rectangle is a square, (3) the line segments joining the midpoints of opposite sidesof any quadrilateral bisect each other, (4) the diagonals of a rectangle are equal in length, (5) the diagonalsof a rhombus are perpendicular, etc.

59

60 10 The Cartesian plane

Problems

10.1. What is the slope of the line joining A and B?

B

A

What is the equation of the line? Give the equation of the line perpendicular to the line through A and Bthat passes through (1,2). Find the intersection point of the two lines.

10.2. Use coordinates to prove that the altitudes of a triangle meet at one point.

10.3. Show that the midline of a trapezoid is parallel to the base and its length equals one-half of the sumof the lengths of the bases.

10.4. Show that if the diagonals of a rectangle are perpendicular, then the rectangle is a square.

10.5. Show that the line segments joining the midpoints of opposite sides of any quadrilateral bisect eachother.

10.6. Show that the diagonals of a rectangle bisect each other.

10.7. Show that the diagonals of a rhombus are perpendicular.

10.8. Let A = (a,a+1) and B = (3a+5,a−1).(a) Find the equation of the line through A and B.(b) Find the equation of the line perpendicular to

←→AB and through A.

(c) Find the equation of the line perpendicular to←→AB and through B.

Chapter 11Conic sections

Conic sections are the curves formed when a plane intersects a double cone like in the picture below. Theintersections are of three types; ellipses, hyperbolas or parabolas

Hyperbola

Ellipse

Parabola

Note that the base (or summit) of the double cone is a circle, so we could consider a circle to be a fourthtype of conic section, but a circle is just a particular type of ellipse.

Now we know why these curves are called conic sections. However, this is not the way we usually thinkabout them. We want to study these curves in a coordinate plane so we can find their equations, just as wedid previously with lines.

Definition 11.1. A circle is a set of points of the Cartesian plane that are at the same distance r from a pointC. The number r is called the radius of the circle and C is called the center of the circle.

r

(h,k)

Circle with radius r and center (h,k).

Since every point (x,y) on the circle above is at distance r from (h,k), then√(x−h)2 +(y− k)2 = r

which by squaring both side yields what is known as the equation of that circle.

61

62 11 Conic sections

(x−h)2 +(y− k)2 = r2

Remark 11.1. Note that after foiling the equation of a circle we get that the terms with x2 and y2 have thesame coefficient (one) and same sign (positive).

Definition 11.2. An ellipse consists of all points such that the sum of the distances to two fixed points, F1and F2 is constant. The points F1 and F2 are called the foci (the singular is focus) of the ellipse.

Before getting the equation of an ellipse we need to set a little notation:

1. The center of the ellipse is the midpoint of F1F2. The distance from a focus to the center is denoted c.2. The points where the line

←−→F1F2 (called the major axis) intersects the ellipse are also equidistant from the

center. The distance from one of these points to the center is denoted a.3. The perpendicular line to

←−→F1F2 (called the minor axis) at the center of the ellipse intersects the ellipse in

two points that are also equidistant from the center. The distance from one of these points to the centeris denoted b.

The following picture shows a ‘horizontal’ ellipse with center (h,k) and distances a and b on the axes

ab

(h,k)

Note that if the ellipse were ‘vertical’ then we should interchange the letters a and b, as a should alwaysbe the larger of the two distances. These changes are mostly cosmetic, and thus we will focus on finding theequation of a ‘horizontal’ ellipse just like the one in the picture above.

We first notice that if we found the equation of an ellipse centered at (0,0) then we could just translatethis ellipse, and its equation, to another congruent ellipse with center at any other point (see chapter 12).

As an ellipse is the set of points with same sum of distances to the foci, then we consider the two pointsat distance a from the center and see how far it is from the foci. It is easy to see that the distance to one ofthem is a−c and to the farthest is a+c. Hence, the sum of the distances to the foci of all points in an ellipseis always 2a. If we now look at the two points on the ellipse at distance b from the center we can see that itis equidistant to the foci. Moreover, the segment joining a focus and this point (this segment has length a)is the hypothenuse of a right triangle with legs b and c. It follows, using the Pythagorean theorem, that

b2 + c2 = a2

This equality is needed to get the equation of the ellipse. In order to find this equation we use thedefinition of an ellipse, and we use that the foci are at distance c from (0,0) (the center of the ellipse) andthe ellipse is ‘horizontal’ to get that the foci are (±c,0). It follows that any point on the ellipse satisfies

d((x,y),(−c,0))+d((x,y),(c,0)) = 2a

or, in other words √(x+ c)2 + y2 +

√(x− c)2 + y2 = 2a

After quite a bit of algebra (see example 11.1) we obtain

x2

a2 +y2

b2 = 1

which, when translated to an ellipse centered at (h,k) yields

11 Conic sections 63

(x−h)2

a2 +(y− k)2

b2 = 1

which is what it is known as the equation of the ellipse.Don’t forget that all the work above is for a ‘horizontal’ ellipse. See problem 11.6.

Example 11.1. Let us find the equation of the ellipse with foci (1,1) and (1,5) and sum of distances to thefoci equal to 10

We set the distance equation we discussed above.

d((x,y),(1,1))+d((x,y),(1,5)) = 10

which implies √(x−1)2 +(y−1)2 +

√(x−1)2 +(y−5)2 = 10

we move one of the radicals to the right hand side to get√(x−1)2 +(y−1)2 = 10−

√(x−1)2 +(y−5)2

Now we square both sides and simplify

(x−1)2 +(y−1)2 = 102−20√

(x−1)2 +(y−5)2 +(x−1)2 +(y−5)2

(y−1)2− (y−5)2−100 =−20√

(x−1)2 +(y−5)2

8y−124 =−20√(x−1)2 +(y−5)2

we simplify a little more and we square

(2y−31)2 = 25((x−1)2 +(y−5)2)

Now we just have to clean this up a little, completing the squares on the way,

25(x2−2x+4)+21(y2−6y+9) = 600

Hence,(x−2)2

24+

(y−3)2

200/7= 1

Note that, in this case, the center is (2,3), a =√

24 and b =√

200/7.

Remark 11.2. Note that after foiling the equation of an ellipse we get that the terms with x2 and y2 havedistinct coefficients (if they were the same then we get a circle) but with the same sign (positive).

Definition 11.3. A hyperbola is the set of points such that the difference of the distances to two fixed points,called F1 and F2, is constant. These fixed points are called the foci of the hyperbola.

Before getting the equation of a hyperbola we need to set a little notation:

1. The center of the hyperbola is the midpoint of F1F2. The distance from a focus to the center is denoted c.2. The points where the line

←−→F1F2 intersects the hyperbola, called the vertices of the hyperbola, are also

equidistant from the center. The distance from one of these points to the center is denoted a.It is easy to see that the difference of the distances from a vertex to the foci is going to have to be 2a.Hence, the difference of distances to the foci that is the same for all points on the hyperbola must be 2a.

64 11 Conic sections

3. There are two lines (called the asymptotes of the hyperbola) that set the limits for the graph of thehyperbola. The hyperbola will get as close as one wants from these lines but will never touch or crossthem.

4. We could draw a rectangle with vertices on the asymptotes and that is tangent to the hyperbola at thevertices. This rectangle has base 2a and height 2b. This number b will be important later.

The following picture shows a ‘horizontal’ hyperbola (in red) with center (h,k), distance to the verticesa. It also features the asymptotes (in blue), and the rectangle that shows how to find 2b

2b(h,k)

2a

Note that this picture gives us enough information to find the equations of the asymptotes. In particularthe slopes of these lines are ± b

a .By doing the same type of argument used to find the equation of the ellipse we can find the equation of

a ‘horizontal’ hyperbola centered at (h,k) and distances a, and b as described above:

(x−h)2

a2 − (y− k)2

b2 = 1

Moreover, it is possible to show that a2 +b2 = c2.It is important to mention that for all this I am considering the hyperbola to open East and West. If you

are working with a hyperbola that opens North and South, then you should look at it sidewise to figurewhere the foci will be.

Remark 11.3. Note that after foiling the equation of the hyperbola we get that the terms with x2 and y2

have different signs.

Definition 11.4. A parabola is the set of points that are equidistant from a given point F and a given line `.The point is called the focus of the parabola, and the line is called its directrix.

Before getting the equation of a parabola we need to set a little notation:

1. Let P be the point on ` such that d(P,F) = d(P, `). The midpoint of PF is called V , and it is the vertexof the parabola.

2. The distance from the vertex to the directrix is denoted p. Clearly the distance between F and V is alsop.

3. The line←→V F is called the axis of the parabola. It is perpendicular to the directrix.

In this book we will only consider horizontal or vertical axis and, just like we have done before, we willonly find the equation of one of these cases, and the other equation will be left as an exercise.

Consider a parabola with vertex V = (h,k) and horizontal directrix (and thus a a vertical axis) anddistance p as above. The equation of this curve is

4p(y− k) = (x−h)2

11 Conic sections 65

Note that the distance p might need to be considered to be negative in case the parabola opens downwardsinstead of upwards.

Remark 11.4. The general equation of a conic section (of the types we are considering) will have the form

Ax2 +By2 +Dx+Ey+F = 0

Note that just by looking at the coefficients with x2 and y2 and their signs we can tell what type of conic theequation will yield.

Problems

11.1. Given the equation 4x2 +9y2 = 36. Find the standard form of this ellipse’s equation. Then find thecoordinates of the foci and sketch the graph of the ellipse.

11.2. Repeat the previous problem with 4y2 +9x2 = 36.

11.3. Consider a pool in the shape of a rectangle with a semicircle on the side, as in the picture below.

r

w

Find the maximum area for a pool that must have perimeter equal to 18 meters.

11.4. Find the foci of the ellipse(x−1)2

5+

(y+1)2

3= 1.

11.5. Find the equation of a circle having a diameter with endpoints (1,1) and (5,4).

11.6. Find the equation of a ‘vertical’ ellipse. Find the relation between a, b and c. Repeat the problem witha ‘vertical’ hyperbola.

11.7. Find the equation of a parabola with horizontal axis.

11.8. If the coordinates of the extremes of the diameter of a circle are (3,1) and (6,5), what is the equationof the circle?

11.9. Find the equation of an ellipse that passes through the points (−1,2),(5,2),(2,3), and (2,1).

Chapter 12Transformations

In this chapter we will learn about movements of the plane and what they do to points and lines living onit. These movements will be called transformations, and they will be though of as functions. Specifically, atransformation of the plane is a bijective function from the plane onto the plane.

Definition 12.1. An isometry of the plane is a transformation of the plane that moves everything on theplane simultaneously and that preserves distances.

Equivalently, an isometry is a bijective function φ : E2→ E2 such that

d(x,y) = d(φ(x),φ(y))

Example 12.1. For example, the shifting of the plane one unit to the right is an isometry. Also, a rotation in30◦ around the origin is also an isometry.

Note that if we first rotate in 30◦ and then we shift we get another isometry. Similarly, if we first shift,then rotate and then shift again, and then we rotate twice, then we get an isometry.

Remark 12.1. A transformation that preserves distances will also preserve angles.We can re-define congruence of objects as A∼= B if there is an isometry mapping A to B bijectively. Thus,

the transformation does the ‘we move A over B’ part in our intuitive definition of congruency.

Definition 12.2. The basic isometries of the plane are

1. The shifting of the plane in any (straight line) direction is called a translation. If a shifting sends theorigin to a point P, then we denote this translation as TP.

2. A (counterclockwise) rotation in an angle θ around a point P will be called Rθ ,P, or just Rθ when it isclear what the point P is.

3. A reflection is the transformation that consist in using a line as a mirror to reflect what is in one side ofit to its other side, and vice-versa. We will say that the reflection is across the line `, and we will denotethis as J`.

Remark 12.2. A translation fixes no points, i.e it moves all the points of the plane, a rotation fixes exactlyone point (the point P), and a reflection fixes the points of the line ` and nothing else.

As mentioned in example 12.1, isometries can be used in succession. Since isometries are functions, thenusing an isometry after the other means that the corresponding functions have been composed. In fact, if φ

and ψ are two isometries of the types listed above then performing φ followed by ψ always yields anotherisometry (that might not be of the type described above). For example if we follow a reflection across thex-axis by a translation that is parallel to the x-axis we obtain an isometry that is not a reflection, translationor rotation. This type of isometry is called a glide reflections.

Also, a rotation with center not equal to the origin can be written as a composition of translations anda rotation with center the origin. Similarly, a reflection can be written as a composition of translations,rotations and a reflection across the x-axis (or the y-axis if you please).

67

68 12 Transformations

Remark 12.3. Every isometry of the plane can be constructed by a succession of rotations, translations,and reflections.

In fact, reflections are enough, as two reflections across parallel lines yield a translation, and two reflec-tions across non-parallel lines yield a rotation.

The three basic isometries can be represented by formulas in the Cartesian plane.

1. A translation TP, where P = (a,b) is given by

TP(x,y) = (x+a,y+b)

2. A rotation Rθ with center the origin is given by

Rθ (x,y) = (xcosθ − ysinθ ,xsinθ + ycosθ)

3. A reflection across the x-axis is given by

J(x,y) = (x,−y)

A reflection across the y-axis is given by

J(x,y) = (−x,y)

Definition 12.3. A dilation is a type of transformation of the plane that does not preserve distances (andthus it is not a n isometry). Actually, a dilation scales up or down the figures on the plane and preserves thefigure’s angles.

Recall that to scale up/down (or zoom in/out) means that we will scale lengths, and that the effect onareas and volumes is a scaling that is different from the one used for lengths.

Remark 12.4. We can now re-define similarity of figures as A∼ B if there is a dilation (maybe mixed withsome isometries) that maps A onto B bijectively.

A dilation can be represented as a function. For example, if the dilation D reduces shapes to 50% of theirsize, then

D(x,y) =12(x,y) =

(12

x,12

y)

The dilation D that triples the shapes is given by

D(x,y) = 3(x,y) = (3x,3y)

Also, a transformation could dilate the plane under a different factor in the horizontal and vertical direc-tions. For instance,

D(x,y) =(

3x,12

y)

Problems

12.1. In the following picture, the shaded rhombus on the right has been translated to the unshaded rhombuson the left. What is the vector used for this translation?

12 Transformations 69

2

4

6

!2!3 !1

2

12.2. In the Cartesian plane there is a triangle ABC, where A = (−3,2),B = (−1,4) and C = (−2,6). Byusing a reflection across the x-axis we obtain a triangle A′B′C′. Give the coordinates of the vertices of thistriangle.

12.3. (a) Find the ‘formula’ that describes a reflection across the line y = a, for some fixed real number a.(b) Find the ‘formula’ that describes a reflection across the line x = b, for some fixed real number b.

12.4. (a) Compose a reflection across the x-axis with a reflection across the y-axis. What do you get?(b) Compose the isometry found in (a) with a rotation in 45◦. What do you get?(c) Compose a reflection across the x-axis with a reflection across the line y = 1. What do you get?

Chapter 13Probability

In order to find the probability of an event to occur we need to be aware of all the possibilities of the activitywe are looking at. For example, if we want to know the probability of getting tails when flipping a coin,then all possible outcomes are “heads” and “tails”. If we are throwing a dice then the possible end resultsare “1”, “2”, “3”, “4”, “5”, and “6”.

Of course, it is possible that if I throw a dice a gust of wind could blow it away and keep it floatingforever in front of my eyes... but that is not a case that we will consider as a ‘possible outcome’.

13.1 Simple probability

Whenever we want to compute the probability of something we will restrict our attention to the activity wehave to observe to determine whether something occurs. This activity is called an experiment, and all thepossible combinations of outcomes of the experiment are called events. Thus, we will always compute theprobability of an event.

Definition 13.1. The probability of the event X to occur when we perform the experiment E is given by

P(X) =Number of favorable outcomes

Total number of outcomes

where an outcome is favorable if its occurrence means that X occurs.

Example 13.1. The probability of throwing a dice and getting a 3 is 1/6 because there are 6 outcomes andonly one of them (getting a 3) is favorable.

Similarly, the probability of throwing a dice and getting a prime number is 3/6 because out of the 6possible outcomes, exactly three of them are favorable; they are ‘getting a 2’, ‘getting a 3’ and ‘getting a 5’.

Example 13.2. Assume you have a 52-card deck. The probability of choosing one card at random from thedeck and get an ace is 4/52, as getting any of the cards is an outcome (52 in total), but only four of thosecards are aces. Hence, there are exactly 4 favorable outcomes.

Note that, since the number of favorable outcomes, for any experiment and event, will always be at mostthe total number of outcomes, then the probability of an event is always less or equal to 1. Moreover, if theevent is impossible then its probability is 0 and if it is certain then its probability is 1.

Sometimes, counting the outcomes of an experiment is not possible. For example, if the experimentis to throw a javelin at the Olympics and the outcomes are all the possible distances the javelin travelsbefore landing, then the outcomes are ALL real numbers between 0 and 100 meters (the WR is almost 100meters)... this is infinitely many outcomes!

71

72 13 Probability

Example 13.3. Assuming all distances are equally probable to be reached. What is the probability of throw-ing a javelin and land it beyond 75 meters?

The total ‘number’ of outcomes in this case is represented by the 100 meters (from 0 to 100) where thejavelin could land. The favorable outcomes are the last 25 meters (beyond the 75-meter mark) where wewant to land the javelin. It follows that the probability is

P(X) =25

100=

14

Example 13.4. My bag of markers has markers of many different colors: 12 of them are red, 7 are blue, 3are purple, 20 black, and 1 is orange. What is the probability of, without looking, choosing a red marker?

13.2 Probability with multiple events

Let X and Y be two events, we know how to find the probability of them (previous section). Now we wantto compute the probability of the events X-OR-Y , and X-AND-Y .

Note that if X is to throw a fair dice and get a 3, and Y is to throw a fair dice and get a 6 then X-OR-Y ,which is to throw a dice and get a 3 or a 6, could be re-phrased as to throw a dice and get a multiple of 3.We get,

P(X−OR−Y ) =26=

16+

16= P(X)+P(Y )

Nice!However, if X is to throw a fair dice and get an even number, and Y is to throw a fair dice and get a prime

number then X-OR-Y , which is to throw a dice and get a 2, 3, 4, 5 or 6, does not work as well as before

P(X−OR−Y ) =566= 3

6+

36= P(X)+P(Y )

this is because X and Y have favorable events that are common to both (to get a 2), so we we need to subtractthis repeated favorable event. Thus,

P(X−OR−Y ) =56=

36+

36− 1

6= P(X)+P(Y )−P(common event)

Now we realize that the common event can be phrased as X −AND−Y , as we want both X and Y tooccur simultaneously. With this, we have the following theorem.

Theorem 13.1. P(X−OR−Y ) = P(X)+P(Y )−P(X−AND−Y )

Most of the times, this theorem is phrased in terms of unions and intersections. It is not so hard to seethat the favorable outcomes of X−OR−Y are the union of the favorable outcomes of X union the favorableoutcomes of Y . Similarly, the favorable outcomes of X−AND−Y are the union of the favorable outcomesof X intersection the favorable outcomes of Y . Hence,

Theorem 13.2. P(X ∪Y ) = P(X)+P(Y )−P(X ∩Y )

Example 13.5. What is the probability of randomly picking a card out of a 52-card deck and get an ace ora card that is diamonds.

Using the previous theorem,

P(ace or diamond) = P(ace)+P(diamonds)−P(ace and diamond)

Since

13.2 Probability with multiple events 73

P(ace) =452

P(diamonds) =1352

P(ace and diamond) =152

thenP(ace or diamond) =

452

+1352− 1

52=

413∼ 0.3

Sometimes computing P(X ∩Y ) is not easy, as the events X and Y could be interlocked. For example,if we want to compute the probability of getting in your socks drawer and randomly picking a black sockand then a white sock. In this situation, the second tim we pick from the drawer then we have already takensomething out of it, and thus the total number of outcomes has changed. In order to straight this up we willneed to introduce the concepts of dependence and independence of events.

Definition 13.2. Two events x and y are said to be dependent if the outcomes of one of them changes if weassume that the other has occurred. In an obvious way independent events are defined.

Since having two events to hold simultaneously could be thought of as a complex event that consists oftwo events, then it is not so surprising to get that the probability of the complex event to occur is the productof the probability of the two simpler events to occur. But, the possibility of having dependent events yieldsthe following theorem.

Theorem 13.3. Let P(Y/X) be the probability of Y occurring assuming that X occurred, then

P(X ∩Y ) = P(X)P(Y/X)

Example 13.6. Assume a drawer contains 7 red socks, 8 black socks and 10 white socks. What is theprobability of randomly picking two socks, one after the other, and get two black socks?

We want to use the formula above, so we need to find P(black) and P(black/black). Since there are 25socks in the drawer, then

P(black) =825

If we now assume that a black sock has been picked then there are only 24 socks left in the drawer, andonly 7 black socks left. So,

P(black/black) =724

It follows thatP(2 black socks) =

825

724

=7

75

Example 13.7. Assume a drawer contains 7 red socks, 8 black socks and 10 white socks. What is theprobability of randomly picking two socks, one after the other, and get two black of the same color?

We first notice that the probability we want to compute can be written as

P(2 red socks OR 2 black socks OR 2 white socks)

Since, there are no common cases in these three situations (can’t pick 2 socks of one color and 2 socksof a different color), then

P(2 red OR 2 black OR 2 white) = P(2 red)+P(2 black)+P(2 white)

We have already computed P(2 black socks) in exercise 13.6. In a similar way we get

P(2 red socks) =7

256

24=

7100

P(2 white socks) =1025

924

=3

20

It follows thatP(2 red OR 2 black OR 2 white) =

7100

+775

+320

=47

150

74 13 Probability

Remark 13.1. The complement of an event X is the event (denoted X) determined by having favorableoutcomes the set of all possible outcomes that are not favorable outcomes of X .

Since X ∪X will cover the set of all possible outcomes, then

1 = P(X ∪X)

and thus P(X) = 1−P(X).

13.3 Counting

We close this chapter with a set of important tools we will need to find complex probabilities. In the exam-ples we have seen before, both the number of favorable outcomes and the total number of outcomes havebeen fairly easy to find. Now we will learn how to count sets that will not be that easy to count. We startwith the fundamental principle of counting.

Theorem 13.4 (Fundamental principle of counting). If a process P consists of k processes P1,P2, ·,Pkperformed back-to-back, and each process Pi can be done in ni distinct ways, then the number of distinctways the process P can be done is

n = n1n2 · · ·nk

Example 13.8. We want to drive from Fresno to Denver. In our road trip we must go through Las Vegas. Ifthere are three different roads connecting Fresno and Las Vegas, and there are five distinct roads from LasVegas to Denver, then how many distinct roads are there from Fresno to Denver?

We consider the trip from Fresno to Denver as a process that consists of two other processes: Fresno-LasVegas and Las Vegas-Denver. Since we know in how many ways these processes can be performed then weknow that there are 3 ·5 = 15 roads joining Fresno and Denver.

Example 13.9. I own three pairs of shoes, two pairs of pants and ten t-shirts. In how many ways can I getdressed? Meaning how many different outfits shoes-pants-shirts can I assemble with what I have in mycloset?

The process of assembling an outfit has three parts, choosing shoes, choosing pants and choosing at-shirt. It follows that the number of possible distinct outfits is 3 ·2 ·10 = 60.

Definition 13.3. Let n be a natural number, then n!, read n factorial is the product

n! = 1 ·2 · · ·n

and, by convention, 0! = 1.

Now we want to see in how many ways we can choose k things out of a set of n things. This processdepends on whether or not it matters the order we choose things. Order is relevant as, for example, it isnot the same to choose three representatives from a class of 30 students, than to choose a president, vice-president, and a treasurer from a class of 30 students. We are, in both cases, choosing three people out of 30but one student being president or treasurer does make a difference, and thus the second case is essentiallydifferent from the first.

If order matters, then choosing k things out of n can be thought of as placing k objects in k (ordered)boxes. As the figure below shows, in the first box, we can place any of the n things, in the second box, wecan place any of the n−1 things that are left, in the third box we can place any of the n−2 things left, etc.it will all stop at the kth box, where we can place any of the n− k+1 remaining things.

n n−1 n−2 · · · · · · n− k+1

13.3 Counting 75

By using theorem 13.4 we now that the number of ways to choose k things out of n, when order mattersis

n ·n−1 · · · · (n− k+1) =(n ·n−1 · · · · (n− k+1))((n− k) · · ·2 ·1)

(n− k) · · ·2 ·1

=n!

(n− k)!

Now, since choosing k things, order mattering, out of k things can be done in k! ways (using formulaabove), then each bunch of k, things with no order mattering, yields k! possible sets of with order mattering.It follows that the number of ways to choose k things out of n, when order does not matter is

n!k!(n− k)!

which is also denoted(n

k

), read ‘n choose k’.

Definition 13.4. A way to choose k distinct things out of n, when order matters, is called a permutation, thetotal number of these permutations is denoted P(n,k). A way to choose k things out of n, when order doesnot matter, is called a combination, the total number of these permutations is denoted C(n,k).

Summarizing.

Theorem 13.5. Let n,k be natural numbers and k ≤ n. Then

C(n,k) =(

nk

)=

n!k!(n− k)!

andP(n,k) =

n!(n− k)!

Example 13.10. What is the probability of choosing four cards off a 52-card deck so that the first is A♦,the second is 2♣, the third is 3♥, and the fourth is 4♠?

In this experiment, the possible outcomes are sets of four cards, chosen in order. It follows that the totalnumber of outcomes is given by all possible ways to choose 4 cards out of the 52 in the deck, IN ORDER.So, the total number of outcomes is

P(52,4) =52!

(52−4)!=

52!48!

= 52 ·51 ·50 ·49 = 6,497,400

The favorable outcomes have to be found in the set of all favorable outcomes. Since we want exactly oneof those cases to occur, the there is exactly one favorable outcome. Hence, the probability asked is

P =1

6,497,400= 0.00000015

which is a 0.000015% chance. Very low.

Example 13.11. What is the probability of choosing four cards off a 52-card deck so that the first is an A,the second is a 2, the third is a 3, and the fourth is a 4?

Note that in this problem the total number of outcomes is equal to the previous example. The favorableoutcomes are different, though. In this case, the first card must be an ace, thus there are four cards that work.For the second, we have four more, either of the 2’s, etc. It follows that the number of favorable outcomesis given by

4 4 4 4

76 13 Probability

which is 44. Hence, the probability asked is

P =44

6,497,400= 0.0000394

which is a 0.00394% chance. Very low, but not as low as in the previous example.

Example 13.12. What is the probability of choosing four cards off a 52-card deck simultaneously so thatthey are A,2,3 and 4?

Since in this case we are choosing four cards simultaneously, then order is not relevant. It follows thatall the outcomes are sets of four cards out of the 52 in the deck, and that order does NOT matter. Hence, thetotal number of outcomes is (using computations done in the previous examples)

C(52,4) =14!

52!(52−4)!

=6,497,400

24= 270,725

The number of favorable outcomes is exactly the same as in the previous exercise. Hence, the probabilityasked is

P =44

270,725= 0.0002364

which is a 0.02364% chance. Very low again.

Problems

13.1. I throw a fair dice with 20 sides (a d20). What is the probability of getting a prime number?

13.2. My bag of markers has markers of many different colors: 12 of them are red, 7 are blue, 3 are purple,20 black, and 1 is orange. What is the probability of, without looking, choosing a red marker?

13.3. Consider a square with side 10 and its inscribed circle. What is the probability that if I randomlychoose a point ‘inside’ the square, then the point is inside the circle?

13.4. Assume that the same number of people are born every day of the year (365 days). What is theprobability of choosing somebody on the street at random and have that person’s birthday to be in a monthending in “ary”.

13.5. This is a classical problem, excellent to try in class.What is the probability of two people in a class of 23 students to have the same birthday?

Chapter 14Statistics

Statistics can be defined as the science that is used to analyze data. For many this is a subarea of mathematicsand for many others it is a completely different science. We will start with simple data analysis and later wewill see how to use data to test assumptions.

14.1 Analysis of data

In order to study data (in a statistical way) we need them to be expressed in terms of numbers. We cannotfind, for example, the mean of the colors of the rainbow, as colors cannot be added, multiplied, etc. Hence,whenever data is given to us we will put them in a list x1,x2, · · · ,xn, where all the xi’s are real numbers. Notethat some of these xi’s may appear more than once in the list, and that since we are dealing with numbershere then we can always re-arrange the data in an increasing (or decreasing) order.

Definition 14.1. The number of times a value xi appears in the list of data is called the frequency of xi.Many times, the frequency of xi is denoted fi.

We want to learn about how the data obtained is distributed. Most of the times we want this to be able tomake decisions. For instance, consider the following data

−1,−1,0,1,1,1,3,3,3,3,3,5,5,5,5,5,5,5,5,6,6

and assume it gives how early/late you arrive to the classroom (negative being early). Then by looking at thedistribution of the data we know that you most of the times are getting at least three minutes late to class.

This way to present data is fine and good but it is not very easy to use when there are many valuesinvolved. In order to simplify this we will use ordered pairs (xi, fi), where xi is a datum and fi its frequency.In this case we consider the data given with no repetitions because each datum will be attached to itsfrequency. For instance, the following data given above can also be written as

(−1,2),(0,1),(1,3),(3,5),(5,8),(6,2)

Now it is obvious that these points can be put in the Cartesian plane ( see chapter 10), where the x-axisis for the data values and the y-axis is for the frequencies. Note that these points could even be consideredas points on a curve, as in the following picture, which is a graphic presentation of how the data wasdistributed.

77

78 14 Statistics

i

7 8!2 !1 x

f

i

6

4

5

3

2

1

7

8

1 2 3 4 5 6

Most of the times these graphs will have a ‘bell-shape’, that could be skew to one of the sides as thefollowing pictures show.

Symmetric Skewed to the left Skewed to the right

The distribution of data given by a graph that looks like the symmetric graph above is called a normaldistribution. We will later discuss these graphs again.

Let us start looking at ways to decide how the data is distributed. The main tools will be to determine whatvalue ‘represents’ better the data given. We will define the three most used measures of central tendencynext.

1. The arithmetic mean of x1,x2, · · · ,xn is

x =x1 + x2 + · · ·+ xn

n

Some times the mean is denoted µ , this is mostly when the data is about population. Also, note thatGPA’s is usually the arithmetic mean of the student’s grades.

2. The mode of x1,x2, · · · ,xn is the value that has the highest frequency. In case there are many values withhighest frequency then all of them (or each one of them, if you prefer) are the mode of that sample ofdata.

3. The median of x1,x2, · · · ,xn is the value that is right in the middle of our increasing (or decreasing) listof values. In case there is no middle value (there are an even number of values) then the arithmetic meanof the two middle values is considered to be the median of the sample.

Example 14.1. Consider the data used in a previous example

−1,−1,0,1,1,1,3,3,3,3,3,5,5,5,5,5,5,5,5,6,6

Then,

x =−1+−1+0+1+1+1+3+3+3+3+3+5+5+5+5+5+5+5+5+6+6

21

=6821

∼ 3.24

14.1 Analysis of data 79

The mode is clearly 5 (it has frequency 8), and the median is 3.

I don’t know what you think about computing arithmetic means but it seems to be too long, and it couldget really long when the number of values given increases. So, note that

x =−1+−1+0+1+1+1+3+3+3+3+3+5+5+5+5+5+5+5+5+6+6

21

=−1(2)+0(1)+1(3)+3(5)+5(8)+6(2)

21

So, the arithmetic mean of x1,x2, · · · ,xn can also be computed by re-writing the data using ordered pairs(xi, fi) and then the numerator of x is obtained by adding all the products xi fi for distinct values of xi. Thedenominator or x stays the total number of values in the sample.

Example 14.2. Consider two companies, A and B, each with 21 people working in them. In company Aeverybody makes $ 100,000 a year, and in company B the owner makes $ 1,000,000 a year, ten employeesmake $100,000 a year and other ten make $ 10,000 a year.

The arithmetic mean of the salaries in company A is

100,000(21)21

= 100,000

per year.The arithmetic mean of the salaries in company B is

1,000,000(1)+100,000(10)+10,000(10)21

= 100,000

per year.So, the arithmetic mean does not catch the difference of salaries between the two companies.The median for company A and B is the same ($ 100,000 a year). The mode in company A is $ 100,000

a year, and the mode in company B is both $ 100,000 a year and $ 10,000 a year. Here some difference wascaught but it would be very easy to manipulate the info to say that the mode of the salaries in company B isjust $ 100,000 a year.... in which case these two companies would have no noticeable difference, but thereis a big difference!!!

It is pretty clear now that we need to add ways to discriminate between data given to us, beyond lookingat the mean, mode and median. We need to know how spread the data is. Note that in the previous example,one of the companies had all the data at one value (same salary), the other one had the data more spread,with one salary being very far from all the others. We will use the standard deviation and the variance asmeasures of dispersion.

Definition 14.2. The standard deviation of x1,x2, · · · ,xn is

S =

√∑(xi− x)2

n

this can also be written as

S =

√(x1− x)2 +(x2− x)2 + · · ·+(xn− x)2

nSometimes, the letter σ is used to denote the standard deviation, but this is mostly used when the data is

related to population.

Since the square root in the previous definition is a little too ugly, then we consider the following concept

Definition 14.3. The variance of x1,x2, · · · ,xn is the square of the standard deviation of the same data.

80 14 Statistics

Example 14.3. Consider the data given in example 14.2. Let us find the standard deviation of the salariesof both companies.

For company A, the mean is x = 100,000, and every value considered is also 100,000, it follows thatxi− x = 0 for all i, and thus SA = 0.

For company B, the mean is x = 100,000, and many values are equal to 100,000, but not all of them. So,we need to actually compute the variance

S2B =

(1,000,000−100,000)2 +(100,000−100,000)2(10)+(10,000−100,000)2(10)21

=(900,000)2 +0+(−90,000)2(10)

21

=(900,000)2 +(90,000)2(10)

21∼ 42428571430

and thus SB ∼ 205,981.9687.

So, the standard deviation does capture that the salaries are not distributed in the same way for the twocompanies. In fact, most of the times the three measures of central tendency and two measures of dispersionwe have learned about in this chapter give a pretty good idea on how the data is distributed. Trust them,these are the main tools you will need to understand and know how to use for your test!

Let us recall that the data given, when written as a set of ordered pairs (datum, frequency), could forma graph that most of the times looks like a bell, or a skewed one (do not assume these graphs always looklike this). Now think of a vertical line that cuts the graph (not necessarily a bell) into two regions (red linein the picture below)

p

30 %70 %

so that 70% of the area under the graph is to the left of the line, and of course 30% is to its right. Thisline is associated to a datum (labeled p in the picture). This value is called the 70th percentile of the datarepresented by the graph.

More precisely, the kth percentile of a set of numbers arranged in an increasing order is the value thathas k% of the numbers below it... and (100−k)% above it. When the area below the graph (not necessarilya bell) is partitioned into four equally big parts by three values (associated to the three red lines)

25%25%25% 25%

then we have found the three quartiles of the data represented by the graph. These values are denoted lowerquartile, median quartile and upper quartile (from left to right). In other words, the lower quartile is the 25th

percentile, the median quartile is the 50th percentile, and the upper quartile is the 75th percentile.

One of the reasons that all the pictures above have been bells, is because if the data given to us is normallydistributed, then we can learn quite a few things about them For instance, if data is in a normal distribution,

14.2 Hypothesis testing 81

then the graph that represents the data is symmetric and this forces the mean, mode and median to be thesame value, which is the 50th percentile. Moreover, in this case the standard deviation S gives very important(although approximate) information about how the data is distributed. Specifically, about 68% of the datawill be at distance at most S from the mean x, about 95% of the data will be at distance at most 2S from x,and about 99.7% of the data will be at distance at most 3S from x. We can see this in the following picture

x

S S S S S S

99.7%

68%95%

Fig. 14.1 Data distribution and distance to the mean.

14.2 Hypothesis testing

In this section we will learn how to use statistics to investigate the validity of an assumption by studyingdata that is related to such assumption. Most of the times we will have a fresh set of data that needs to becompared with old data, or we will have data that we would like to compare with theoretical values.

We will call null hypothesis to the hypothesis/assumption/statement that implies no change. So, the nullhypothesis should always include the word ‘equal’ (or the equal sign). The final decision we will make willbe either that the null hypothesis is rejected or that there is no sufficient evidence to reject it.

The negation of the null hypothesis is called the alternative hypothesis. We will denote the null hypoth-esis by H0 and the alternative hypothesis by H1.

Example 14.4. Assume that the scores in a math class (CSET II, for instance) ten years ago were 43% gotA’s, 29% were B’s, 14% were C’s and there was also a 14% of D’s.

Last year out of 500 people taking the test, there were 199 A’s, 153 B’s, 70 C’s and 78 D’s.We want to show that the current distribution of the data is different from that one ten years ago. Hence,H0 = the current data distribution is the same as the distribution ten years ago.H1 = the current data distribution differs from the distribution ten years ago.Now note that the mean for the set of old data cannot be computed as we have done before because we

do not know how many values were considered to find those percentages. So, what we do is that we let 100to be the total number of data, and the percentages will dictate the frequencies of the values. Also, since wecannot compute the mean of letters then we need to transform the grades into numbers (we use the standardscale from 1 to 4 for this). Hence,

xold =43 ·4+29 ·3+14 ·2+14 ·1

100

=172+87+28+14

100

=301100

= 3.01

82 14 Statistics

xnew =199 ·4+153 ·3+70 ·2+78 ·1

500

=796+459+140+78

500

=1473500

= 2.946

We can see that the means are fairly close, but recall that sometimes the mean is not enough to determinethe distribution of data. So, let is compute the standard deviations.

S2old =

43 · (4−3.01)2 +29 · (3−3.01)2 +14 · (2−3.01)2 +14 · (1−3.01)2

100

=43 · (0.99)2 +29 · (−0.01)2 +14 · (−1.01)2 +14 · (1−2.01)2

100

=42.1443+0.0029+14.2814+56.5614

100= 1.1299

So, Sold =√

1.1299∼ 1.063

S2new =

199 · (4−2.946)2 +153 · (3−2.946)2 +70 · (2−2.946)2 +78 · (1−2.946)2

500

=199 · (1.054)2 +153 · (0.054)2 +70 · (−0.946)2 +78 · (−1.946)2

500

=221.07228+0.446148+62.64412+295.37945

500= 1.159084

So, Snew =√

1.159084∼ 1.076.Just like with the means, the standard deviations are pretty close. But, let us not rush any judgement.

We now know about the null hypotheses but how can we test it? Since data can come in different amountsand orders of magnitude, we will need to standardize the values given to us. By doing this we will be able todisplay the data in a very generic way, and thus we will be able to use pre-constructed tables (see appendix)that will do most of the computational work for us!!

The idea is to take old data (or theoretical data) and compare it with new data. We will do this by usingtwo types of tests: one-sample z-test and one-sample χ2-test.

In both tests, we will reject the null hypothesis if the test score that represents our data falls farther thana certain critical value. This critical value depends on the test used and on the information given in theproblem.

z-score

For the one-sample z-score we will assume our data (new and old) to be distributed normally (bell shape).In this case we will use a formula that will standardize the data and give us a number (called the z-score).This score will tell us whether our data is ‘too far away’ from the (standardize) old mean, this will be doneby just comparing the score obtained with a value associated to the percentage of significance we want togive to the test.

Example 14.5. Assume the z-score of the sample is z= 2.34 then the new data is 2.34 (standardize) standarddeviations from the (standardize) old mean. Now we look at table 1 in the appendix and see that for z = 2.34the area (on the right half of the normal bell curve) is given by the table as A = .49. This means that the

14.2 Hypothesis testing 83

new data is in the farthest 100− 2 · 49 = 2% of the old data, that is pretty far away! (most of the timesthe acceptable percentage will be 5%). Hence, the null hypothesis is rejected, which most of the times isconsidered good (things have stayed the same).

The z-score will allow us to compare two sets of data. As mentioned before, in most cases, one set is our‘control’ set; either known data from the past or data that is generated out of theoretical work. The secondset is, most of the times, recent data obtained from some type of poll, study, etc.

Assume you know the following:(i) x = new data mean,(ii) µ = theoretical mean or old mean,(iii) s = new data standard deviation,(iv) n = number of new data.then the z-score is

z =x−µ

s /√

n

where the value s√n

is called the standard error of the (new data) mean.The z-score obtained can then be looked in table 1 in the appendix to see what we can say about the null

hypothesis. Note that a z-score of 1.96 yields an area under one side of the bell of .475, which means an areaof 2 ·0.475 = .95 overall (a 95% of it all), that means that any z-score larger than 1.96 (or less that −1.96)implies that our data falls below the 5% of significance, and thus the null hypothesis would be rejected.

Example 14.6. With the same data of example 14.4. We want to show that the new scores are statisticallydifferent from the old ones with a 5% of significance. So, we need to take H0 to be ‘the current data distri-bution is the same as the distribution ten years ago’ (note that word equal, although not used, is implied).

Note that we are considering two situations at the same time here: the new scores could be higher orlower than the old ones.

From example 14.4 we know(i) x = 2.946 (new data mean)(ii) µ = 3.01 (old data mean)(iii) s = 1.076 (new standard deviation)(iv) n = 500 (number of new data)then the standard error is

1.076√500∼ 0.048

and thus the z-score is

z =2.946−3.01

0.048∼−1.33

We look at table 1 in the appendix and we see that z = 1.33 corresponds to (one-sided) area A = .408,which means a double-sided area of .816, which is much closer than the .95 that the 5% of significance asksfor. It follows that the null hypothesis cannot be rejected, and thus we can say that the new data is consistentwith the old one.

Remark 14.1. If the hypothesis we wanted to show true would have been that the new data was ‘larger’than the old one, then, since the z-score has (x− µ) in its numerator then we would have been interestedonly in positive values for the z-score, and thus the area to consider should be only on the right hand side ofthe bell.

84 14 Statistics

χ2-test

This test is similar to the z-score test, but in this one we use a different table (table 2 in the appendix). Asbefore we want to test new data by comparing it with old data (or theoretical data).

We first compute the ‘χ2-score’,

χ2 =

(n−1)s2

σ2

where s2 = new data variance. σ2 = theoretical, or old, variance, and n = number of items in the new dataset.

As usual, the data must be analyzed with certain level of significance, this number must be found in thefirst row of table 2. For instance a 5% of significance will place us at the 0.05 entry (third row). Next wemust find the number of degrees of freedom of our problem. This value is given by n− 1, where n is thenumber of values in the new data set. Once this is known, this number must be located in the extreme leftcolumn of table 2. The intersection of the appropriate row and column will give us the critical value, whichwe will compare to the χ2-score to see if we are able to reject the null hypothesis. Specifically, as we didwith the z-score, if we get a χ2-score that is beyond the critical value, then we will reject the null-hypothesis.

Example 14.7. Just as we did in example 14.6. With the same data of example 14.4. We want to show thatthe new scores are statistically different from the old ones with a 5% of significance. So, we need to takeH0 to be ‘the current data distribution is the same as the distribution ten years ago’.

From example 14.4 we know(i) σ = 1.063 (old standard deviation)(ii) s = 1.076 (new standard deviation)(iii) n = 500 (number of new data)

But this is too many data for our table! So, we will ‘standardize’ them a little. Since there are 500 dataand our table gets to 100 degrees of freedom, then we will divide everything by 5, without really changingthe variances.

Then the χ2-score is

χ2 =

(n−1)s2

σ2 =(100−1)(1.076)2

(1.063)2 = 101.4362553309

Since the critical value corresponding to 5% significance and 99 degrees of freedom is 113.145 (foundin table 2) then our χ2-score is lower than the critical value, and thus we cannot reject the null hypothesis.

14.3 Curve fitting

In order to understand how data behaves, we might want to see how they relate to another set of data, orwith theoretical results. Also, sometimes one wants to see if there is any hint of two results being connected.For instance, we would like to know whether the percentage of population that can swim depends on thedistance from people to the closest beach.

In the previous two situations we wanted to see if there was a relation between two sets of data. In otherwords, we wanted to see if there was a correlation between them. In statistical terms a correlation coefficientis a number that measures how directly close (if at all) two sets of data are to each other. In this case we areusing the word ‘directly’ to mean a proportional relation between the two sets of data, this can be phrasedin other words by saying that if we plotted points (xi,yi), where the xi’s are from one set of data and theyi’s are from the other and then the correlation coefficient would determine how close are these points frombeing on a straight line.

The correlation coefficient will always be a number between−1 and 1, and whenever it is zero will meanthat the tow sets of data are probably not on a linear relation. Positive or negative coefficients would mean

14.3 Curve fitting 85

that the line that approximates the plotted points will have positive or negative slope. This number is foundby computing

r =∑(x− x)(y− y)√

∑(x− x)2 ∑(y− y)2

Moreover, if the regression coefficient is not zero, then the data should be close to be a line. The line thatis the ‘closest’ to the data is called the regression line, and the process to find it is called linear regression.The equation of this line is given by y= a+bx, where y is considered to be depending on x, and the numbersa and b are given by

b =∑xy−nx y∑x2−nx2 a =

∑y−b∑xn

As usual, n denotes the number of data.

Example 14.8. Assume that the following table shows the percentages of people who can swim living in acertain town, and the distance from the town to the nearest beach. We will consider the number of swimmersto depend on the distance to a beach.

distance (miles) % swimmersTown 1 100 56Town 2 0 82Town 3 600 50Town 4 500 70Town 5 60 76Town 6 300 62

Then we have the point (x1,y1) attached to Town i. In order to find the correlation coefficient we need tofind x and y. We get,

x =100+0+600+500+60+300

6= 260

y =56+82+50+70+76+62

6= 66

It follows that

r =−0.607751 a =−0.0298 b = 73.7662

Computations are long and it is very easy to make mistakes doing them. So, we will do them using acalculator (see next section).

Now, what does the regression line y = ax+ b we have just found mean to us? Most of the times it isused to approximate some info about the way the data increases/decreases. For instance, in example 14.8the value of b is the y-intercept of the line, and thus it is the value that should correspond to x = 0, whichcould be interpreted as the percentage of people that should be expected to swim for a town by a beach(distance 0 miles to the closest beach). Note that b ∼ 74 is consistent with the data we have for Town 2.Also, looking at a (the slope of the line) we could say that every mile getting away from a beach shoulddecrease the percentage of swimmers by approximately 0.3%. However, it is important to note that thecorrelation coefficient is r ∼ −0.6, which is not that small, and thus the approximation could be not thatgood for certain values.

Example 14.9. With the data given in example 14.8. What should be the percentage of swimmers living ina town that is 200 miles from the closest beach?

Since the data is approximated by y = −0.0298x+ 73.766, then the number of swimmers, given by y,that corresponds to x = 200 is y =−0.0298 ·200+73.766 = 67.806, which is fairly consistent with the infowe have from Town 1 and Town 6.

86 14 Statistics

14.4 Calculator Use

Most calculator work in a similar way. Out of the ones that you are allowed to use in the CSET, we will goover how to use the TI-83 (or similar) and the HP 9g (or similar).

14.4.1 TI-83

In order to ask a calculator to do the long calculations needed in this test we first need to be able to createlists of data. I am using a TI-83 Plus, slight differences may exist with other calculators.

Press STAT to get to a window that says EDIT - CALC - TEST at the top. Move the cursor if necessaryto get EDIT shaded, then press ENTER . Now you can see lists of data, if they are not empty you candelete their contents by getting the cursor on the header of the list and press CLEAR .

Once the lists are emptied you can input your data. Just move the cursor over L1, for instance, and dropthe cursor to the next cell, add the datum and press ENTER to input the next datum. When you are done

inputing data move your cursor to another list, if you want to create another list, quit by pressing 2nd

QUIT , or just plain start a new (statistics) process without quitting by pressing STAT . Note that most ofthe times when you create lists, you will create lists with the same number of data, also, please be sure thatif you have data that are related, then they are in the same order when they are inputed. For instance, for thedata in example 14.8, you want to see the data in your calculator to be just like it appears on the table givenabove.

When looking for correlation coefficients and/or linear regression, this is a ‘trick’ that makes things alittle easier, as it will give you r, a and b in the same window.

Press 2nd CATALOG (above 0) and move the cursor down until you reach DiagnosticOn, pressENTER twice (you want to get ‘DiagnosticOn Done’).

To start finding r, a and b you need to press STAT , move the cursor to CALC and press 4 (or justpress ENTER after you moved the cursor to LinReg(ax+b)). Now you need to select the lists of data you

want to compare. Press 2nd LIST and then select the list that contains the data for the independent data(the one that will be x in the line you are finding). Now you should be back to the LinReg window, press ,to now add a second list just as we did before, this list is the variable that depends on the data in the previouslist (this data will be represented by y in the line you are finding). Finally press , again and press VARS ,

move the cursor to Y-VARS, press ENTER and select Y1. Now, you are back at the LinReg window for alast time. Press ENTER to get r, a and b.

If at this point you want to graph the data and the regression line you could press Y= and thenGRAPH . In case the data seems not to appear, or it is too small, then you will need to press WINDOW

and carefully input the ranges for the variables x and y so all the data given fits in these ranges. For instance,for the info given in example 14.8 we would like to set Xmin =-100, Xmax=700, the scale to Xscl=100(how far apart the dashes in the x-axis would be), and for the y’s we would want Ymin =0, Ymax=100 andYscl=10.

In order to find the standard deviation, mean, etc of a list of data, you first have to create a list as explainedabove, then you have to press STAT , move the cursor to CALC and choose 1-Var Stats, you should nowbe looking at a window that says “1-Var Stats”. Now you press LIST to choose the list that contains yourdata, select it, and when you get to the next window press ENTER . The mean is x, the standard deviationis σx, and the median is Med.

14.4 Calculator Use 87

14.4.2 HP 9g

This one works slightly different, as you have to make the list of data when you have already chosen whatcomputations you want to make.

In order to find the correlation coefficient and the regression line you need to start by pressing Mode ,then choose 1Stat and press ENTER , at this point you are in statistics mode, and thus non-statistical cal-culations will not work as usual. After you have pressed ENTER you need to select Reg, press ENTER ,select Lin and press ENTER again. Now you need to input the data. Press DATA and select Data Input.Now you will input data in pairs, the x and y that are linked are inputted right after the other with pressingH in between values. For instance, the data in example 14.8 is inputted as

100 H 56 H 0 H 82 H 600 H 50 H 500 H 70 H 60 H 76 H 300 H 62

Do not press H after the last value.

Finally, you press 2nd STATVAR and by placing the cursor on a, b or r you will see their value on thebottom right corner of the screen. The correlation coefficient is r and the regression line is y = bx+a.

88 14 Statistics

Appendix: Tables.

In the following table, the columns under an A shows the area under the bell curve that is between z = 0 andthe (positive) value of z that is by it. Some times this table is called a table of z-scores.

Note that since the bell is symmetric then we will not give the table for the negative values of z.

Table 1: z-scores table

z A z A z A z A z A z A z A

.00 .000 .48 .184 .96 .331 1.44 .425 1.92 .473 2.40 .492 2.88 .498

.01 .004 .49 .188 .97 .334 1.45 .426 1.93 .473 2.41 .492 2.89 .498

.02 .008 .50 .191 .98 .336 1.46 .428 1.94 .474 2.42 .492 2.90 .498

.03 .012 .51 .195 0.99 .339 1.47 .429 1.95 .474 2.43 .492 2.91 .498

.04 .016 .52 .198 1.00 .341 1.48 .431 1.96 .475 2.44 .493 2.92 .498

.05 .020 .53 .202 1.01 .344 1.49 .432 1.97 .476 2.45 .493 2.93 .498

.06 .024 .54 .205 1.02 .346 1.50 .433 1.98 .476 2.46 .493 2.94 .498

.07 .028 .55 .209 1.03 .348 1.51 .434 1.99 .477 2.47 .493 2.95 .498

.08 .032 .56 .212 1.04 .351 1.52 .436 2.00 .477 2.48 .493 2.96 .498

.09 .036 .57 .216 1.05 .353 1.53 .437 2.01 .478 2.49 .494 2.97 .499

.10 .040 .58 .219 1.06 .355 1.54 .438 2.02 .478 2.50 .494 2.98 .499

.11 .044 .59 .222 1.07 .358 1.55 .439 2.03 .479 2.51 .494 2.99 .499

.12 .048 .60 .226 1.08 .360 1.56 .441 2.04 .479 2.52 .494 3.00 .499

.13 .052 .61 .229 1.09 .362 1.57 .442 2.05 .480 2.53 .494 3.01 .499

.14 .056 .62 .232 1.10 .364 1.58 .443 2.06 .480 2.54 .494 3.02 .499

.15 .060 .63 .236 1.11 .367 1.59 .444 2.07 .481 2.55 .495 3.03 .499

.16 .064 .64 .239 1.12 .369 1.60 .445 2.08 .481 2.56 .495 3.04 .499

.17 .067 .65 .242 1.13 .371 1.61 .446 2.09 .482 2.57 .495 3.05 .499

.18 .071 .66 .245 1.14 .373 1.62 .447 2.10 .482 2.58 .495 3.06 .499

.19 .075 .67 .249 1.15 .375 1.63 .448 2.11 .483 2.59 .495 3.07 .499

.20 .079 .68 .252 1.16 .377 1.64 .449 2.12 .483 2.60 .495 3.08 .499

.21 .083 .69 .255 1.17 .379 1.65 .451 2.13 .483 2.61 .495 3.09 .499

.22 .087 .70 .258 1.18 .381 1.66 .452 2.14 .484 2.62 .496 3.10 .499

.23 .091 .71 .261 1.19 .383 1.67 .453 2.15 .484 2.63 .496 3.11 .499

.24 .095 .72 .264 1.20 .385 1.68 .454 2.16 .485 2.64 .496 3.12 .499

.25 .099 .73 .267 1.21 .387 1.69 .454 2.17 .485 2.65 .496 3.13 .499

.26 .103 .74 .270 1.22 .389 1.70 .455 2.18 .485 2.66 .496 3.14 .499

.27 .106 .75 .273 1.23 .391 1.71 .456 2.19 .486 2.67 .496 3.15 .499

.28 .110 .76 .276 1.24 .393 1.72 .457 2.20 .486 2.68 .496 3.16 .499

.29 .114 .77 .279 1.25 .394 1.73 .458 2.21 .486 2.69 .496 3.17 .499

.30 .118 .78 .282 1.26 .396 1.74 .459 2.22 .487 2.70 .497 3.18 .499

.31 .122 .79 .285 1.27 .398 1.75 .460 2.23 .487 2.71 .497 3.19 .499

.32 .126 .80 .288 1.28 .400 1.76 .461 2.24 .487 2.72 .497 3.20 .499

.33 .129 .81 .291 1.29 .401 1.77 .462 2.25 .488 2.73 .497 3.21 .499

.34 .133 .82 .294 1.30 .403 1.78 .462 2.26 .488 2.74 .497 3.22 .499

.35 .137 .83 .297 1.31 .405 1.79 .463 2.27 .488 2.75 .497 3.23 .499

.36 .141 .84 .300 1.32 .407 1.80 .464 2.28 .489 2.76 .497 3.24 .499

.37 .144 .85 .302 1.33 .408 1.81 .465 2.29 .489 2.77 .497 3.25 .499

.38 .148 .86 .305 1.34 .410 1.82 .466 2.30 .489 2.78 .497 3.26 .499

.39 .152 .87 .308 1.35 .411 1.83 .466 2.31 .490 2.79 .497 3.27 .499

.40 .155 .88 .311 1.36 .413 1.84 .467 2.32 .490 2.80 .497 3.28 .499

.41 .159 .89 .313 1.37 .415 1.85 .468 2.33 .490 2.81 .498 3.29 .499

.42 .163 .90 .316 1.38 .416 1.86 .469 2.34 .490 2.82 .498 3.30 .500

.43 .166 .91 .319 1.39 .418 1.87 .469 2.35 .491 2.83 .498 3.31 .500

.44 .170 .92 .321 1.40 .419 1.88 .470 2.36 .491 2.84 .498 3.32 .500

.45 .174 .93 .324 1.41 .421 1.89 .471 2.37 .491 2.85 .498 3.33 .500

.46 .177 .94 .326 1.42 .422 1.90 .471 2.38 .491 2.86 .498 3.34 .500

.47 .181 .95 .329 1.43 .424 1.91 .472 2.39 .492 2.87 .498 3.35 .500

14.4 Calculator Use 89

In the following table the χ2 values corresponding to the probability of exceeding the critical value, orlevel of significance (top row) and the degrees of freedom (left column).

Table 2: χ2-test table

4/13/10 1:14 PMTable: Chi-Square Probabilities

Page 1 of 1http://people.richland.edu/james/lecture/m170/tbl-chi.html

Table: Chi-Square Probabilities

The areas given across the top are the areas to the right of the critical value. To look up an area on the left,subtract it from one, and then look it up (ie: 0.05 on the left is 0.95 on the right)

df 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005

1 --- --- 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879

2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597

3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838

4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860

5 0.412 0.554 0.831 1.145 1.610 9.236 11.070 12.833 15.086 16.750

6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548

7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278

8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955

9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589

10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188

11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757

12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.300

13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.688 29.819

14 4.075 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.141 31.319

15 4.601 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.578 32.801

16 5.142 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.000 34.267

17 5.697 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.409 35.718

18 6.265 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.805 37.156

19 6.844 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.191 38.582

20 7.434 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.566 39.997

21 8.034 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.932 41.401

22 8.643 9.542 10.982 12.338 14.041 30.813 33.924 36.781 40.289 42.796

23 9.260 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.638 44.181

24 9.886 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.980 45.559

25 10.520 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.314 46.928

26 11.160 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.642 48.290

27 11.808 12.879 14.573 16.151 18.114 36.741 40.113 43.195 46.963 49.645

28 12.461 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.278 50.993

29 13.121 14.256 16.047 17.708 19.768 39.087 42.557 45.722 49.588 52.336

30 13.787 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.892 53.672

40 20.707 22.164 24.433 26.509 29.051 51.805 55.758 59.342 63.691 66.766

50 27.991 29.707 32.357 34.764 37.689 63.167 67.505 71.420 76.154 79.490

60 35.534 37.485 40.482 43.188 46.459 74.397 79.082 83.298 88.379 91.952

70 43.275 45.442 48.758 51.739 55.329 85.527 90.531 95.023 100.425 104.215

80 51.172 53.540 57.153 60.391 64.278 96.578 101.879 106.629 112.329 116.321

90 59.196 61.754 65.647 69.126 73.291 107.565 113.145 118.136 124.116 128.299

100 67.328 70.065 74.222 77.929 82.358 118.498 124.342 129.561 135.807 140.169

Table of Contents

Problems

14.1. Find the standard deviation, variance, lower quartile, median quartile and upper quartile of the datagiven in example 14.1.

14.2. Explain/prove that if data is in a normal distribution then its mean, mode and median coincide.

14.3. Using the data given in example 14.1, and problem 14.1 find the 2.5th percentile, the 16th percentileand two values so that 81.5% of the data is between them.

Solutions

Chapter 1

1.1 By contradiction. Assume that a triangle has more than one right angle, then the sum of the angles ofthe triangle is going to be more than 180◦, which is impossible! It follows that a triangle can have at mostone right angle.

1.2 A circle centered at C.

1.3 Let AB be a segment of length 5. Draw a circle with radius 2 centered at A and another with radius 4centered at B. Since 2+4 > 5 the these circles will intersect. Take one of the points of intersection and callit C. Join A with C and B with C,4ABC has sides as asked.

1.4 It is impossible because 2+4 < 7 and thus, following the construction in the previous problem, the twocircles drawn will not intersect.

Chapter 2

2.1 We first notice that both P and Q are collinear with A and B, since there is a unique line joining twopoints (A and B in this case) then P and Q must be on the line

←→AB.

Let 2d be the distance between A and B. In the picture below, we can see that d is the distance from P toA and also from Q to A, this forces the distance from P to Q to be zero. Contradiction.

QA BP

2.2 Given a segment AB, let C be its midpoint, and let←→CD and

←→CE be two distinct perpendicular bisectors

of AB, this contradicts the fact that these two lines will have a nonzero angle (remark 2.3) formed by themat C.

2.3 Fix an angle ∠ABC and let−→BD and

−→BE be two distinct angle bisectors of ∠ABC. Since these two rays

share their initial point, then they form an angle, which by remark 2.3 it has positive measure. This is acontradiction, as assuming WLOG that

−→BD is an angle bisector forces that

−→BE is not a bisector of ∠ABC.

2.4 There are three different cases. If the four points are collinear then there is just one line, if three ofthem are collinear and the fourth is not, then there are 4 lines; the one containing three points and the onesfrom the non-collinear point to the three on a line. If at most 2 points are on a line (forming something likea square) then there are 6 lines; the four sides of the ‘square’ and the two diagonals.

There are no other cases, as any two points are always on a line.

91

92 14 Solutions

2.5 The intersection of two rays might be: empty (parallel rays, for example), a ray (in case one ray is‘contained’ in another), a point, or a segment, as in the following pictures.

Their union could be an angle, straight line, a cross-like shape, as in the figures below, but it could alsobe two disjoint rays, or just one ray (in case one of the rays is ‘contained’ in the other).

2.6 The ray−→LK.

2.7 The following pictures show all the possible lines one can draw using those ten points. The first oneshows 15 lines and the second the other 5.

2.8 Let us think that an angle is formed by sweeping, counterclockwise, a ray from its ‘starting’ ray toits ‘final’ ray. Hence, for each ray in the figure there are exactly three positive angles ‘starting’ at that ray(positive and less than 360◦) and ending at one of the other three rays. Since there are four rays, then thereare exactly 3 ·4 = 12 angles (positive and less than 360◦) in the figure.

2.9 0◦ = 0 radians, 30◦ = π/6 radians, 60◦ = π/3 radians, 90◦ = π/2 radians, 270◦ = 3π/2 radians, and360◦ = 2π radians.

2.10 P = 2.5 and Q = 5.

2.11 Let x be the measure of the angle we are looking for. We know that its supplement is 180◦ − x.The information given yields x = 3(180◦− x), which implies x = 135◦. The angle is too large to have acomplement.

2.12 Using the vertical angle theorem we get that the ∠ABC = α +β . Since l is an angle bisector, thenα = β .

Chapter 3

3.1

3.2

3.3 (a) Since C is a point not on←→AB then Playfair’s axiom assures the existence of a line through C that is

parallel to←→AB.

(b) Note that at C we have six angles, the three ‘above’ l1 are equal to β ,γ and α (respectively, from left to

14 Solutions 93

right) because of l1||l2, proposition 29 (once with transversal←→AC and anther with transversal

←→BC), and the

vertical angle theorem. It follows that β +γ +α = 180◦, and thus the angle-sum of the triangle is also 180◦.

3.4 This is an activity you should do, so there is no solution.

3.5 This is an activity you should do, so there is no solution. However, after repeating the activity manytimes you probably will figure that the larger the triangle, the larger the sum of the angles of the triangle. Infact, in a sphere with radius r the following theorem holdsTheorem (Girard). Let4 be a spherical triangle with interior angles α , β and γ , then its area is

A4 = r2 (α +β + γ−π)

Note that this theorem confirms that the larger the angle-sum of the triangle, the larger the area of it.

3.6 Consider the figureC

A B D

It is clear that ∠ABC +∠DBC = 180◦. Since we are in Euclidean geometry, then ∠ABC +∠CAB+∠BCA = 180◦. It follows that ∠CAB+∠BCA = ∠DBC.

On the other hand, if we were working in non-Euclidean geometry then ∠ABC+∠CAB+∠BCA 6= 180◦,and thus ∠CAB+∠BCA 6= ∠DBC.

3.7 The bear is white, as the man’s house must be at the North pole. The North pole and the South pole arethe only places on Earth (a sphere) where one can have an equilateral triangle as described in the problem.

3.8 If a triangle has angle sum equal to 200◦ then the triangle lives in an elliptic geometry, it follows thatthere are no parallel lines. So, the answer is zero.

If a triangle has angle sum equal to 100◦ then the triangle lives in a hyperbolic geometry, it follows thatthere are infinitely many parallel lines to ` through P.

Chapter 4

4.1

4.2

4.3

Chapter 5

5.1

5.2

5.3 For any shape, if one zooms ×2 then the area will quadruple. The polygon need not be regular to solvethis problem.

5.4 Since the length of and L shape is equal to a reversed L (all angles are right!), then the perimeter of thefigure is equal to the perimeter of a 24 by 28 rectangle, which is 2(24+28) = 104.

94 14 Solutions

5.5 Since an octagon has interior angle sum equal to 6 ·180◦, then each interior angle measures

6 ·180◦

8= 3 ·45◦ = 135◦

It follows that an exterior angle measures 180◦−135◦ = 45◦.

5.6 Note that the two hexagons are similar and that the ratio of their sides is 2 : 1 (because the ratio oftheir apothems is 2 : 1). But that implies that the the ratio of their areas is 4 : 1. So, if the area of the smallhexagon is A then the area of the big one is 4A. It follows that the shaded area is 3A.

Now we just need to find A, assuming a is known. The measure of every interior angle in a regularhexagon is 120◦, it follows that the length of the side (say 2s) relates to the apothem by

a = s tan60◦ = s√

3

So, the area of each of the six triangles that form the small hexagon is(2 a√

3

)a

2=

a2√

3=

√3a2

3

and thus the area of the smaller hexagon is

A = 6

√3a2

3= 2√

3a2

Finally, the shaded area is 3A = 6√

3a2.

5.7

5.8

5.9

Chapter 6

6.1 We consider an isosceles triangle4ABC with base angles ∠CAB and ∠CBA with measure 2α . We drawthe angle bisectors to get the following picture.

C

A B

!

!!!

D

Since the base angles of4ABD are congruent, then this triangle is isosceles.

6.2 The picture associated to this problem is

14 Solutions 95

x

6

874Where the dashed lines are given by solar rays. Since the angle formed by these rays are the same, and

both triangles are right then the triangles are similar. It follows that

x74

=68

which implies x = 55.5 f t

6.3 The picture associated to this problem is

50

30

x

Since the triangle is right then using the Pythagorean theorem yields that the dashed line (hypothenuse)is approximately 58.3 f t.

6.4 Since 4ABE is inscribed on the square then both the base and the height of it are 6 cm. Thus the areaof 4ABE is (6 ·6)/2 = 18 cm2. It follows that the area of the region external to the triangle is 6 ·6−18 =18 cm2.

Since we know the triangle formed by all the midlines of a triangle is similar to the original triangle withratio 0.5 then the height and base of the smaller triangle must be 3 cm. Hence, its area is (3 ·3)/2 = 4.5 cm2.

The area of the whole purple region is 18+4.5 = 22.5 cm2.

6.5 It is easy to see that ∠ABE = ∠ACB = 65◦ and that ∠EAB = ∠ACD = ∠CBE = 25◦ . Thus4ACD∼=4CAB by ASA. Also,4ACD∼4CAB∼4BAE ∼4CBE.

6.6

6.7 From the picture

c

b a

a

b

a b

a

b

cc

c

we see that the square-like figure in the middle has four congruent sides (length c) and has four rightangles. Thus it is a square.

Now we compute the area in two different ways. first realizing that the figure is a square with side a+b,

and thus the area is (a+b)2. Then we compute the area by adding c2 (the square in the middle) plus 4ab2

(the four triangles), we get

96 14 Solutions

4ab2

+ c2 = (a+b)2

After some algebra we get a2 +b2 = c2.Now with the picture

b

c

a

cc

c

a

bb

b

a

a

we act similarly and we see that the square in the middle has side b−a. Thus computing the areas yields

c2 = 4ab2

+(a−b)2

After some algebra we get a2 +b2 = c2.

6.8 It does not, consider the following figure, where the circle passing through D and C is centered at B

A B

D

C

Note that ∠CAB = ∠DAB is an angle of both 4CAB =4DAB, AB is a side of both 4CAB =4DAB,and BD∼= BC because they are radii of the same circle. Then, there is a correspondence between the verticesof4CAB and4DAB so that we get congruent Angle-Side-Side but the triangles are clearly not congruent.

6.9 No, if the side with length 7 in is taken as the base of the triangle, then the other two sides will nevermeet, as the sum of their lengths is less than 7 in.

6.10 This follows directly from SSS.

6.11

6.12

6.13

6.14

6.15

6.16

14 Solutions 97

Chapter 7

7.1 Since we want to cover the field with no overhang then the width of the square must divide both 360and 200. Now, gcd(360,200) = 40 then we need squares that have width at most 9 f t and that divide 40.So, the largest the squares can be is 8 f t × 8 f t.

7.2 Since the diagonals in a rhombus bisect, then after tracing them we get four triangles, the sides of therhombus are the hypothenuses of these triangles. So, using the Pythagorean theorem we get that each sideis 15 in, and thus the perimeter is 60 in.

7.3 Since we know all the sides’ lengths and that the trapezoid is isosceles then we get the following picture

7

25 25

8

87where the dashed lines are the heights we want to measure. Note that each height can be though of as

a leg of a right triangle with other leg measuring 7 in and hypothenuse measuring 25 in. It follows, by thePythagorean theorem that h = 24 in.

7.4 Since the trapezoid is isosceles, then RS ∼= TW . It follows that 4SRW ∼= 4TWR by SAS. Then,∠T RW ∼= ∠SWR, and thus4RPW is isosceles (congruent base angles).

7.5 Since LP∼= MP then4LPM is isosceles, and thus ∠PLM ∼= ∠PML. But since AD||LM then

∠LPM ∼= ∠PLM ∼= ∠PML∠MPD

Now we use that AP∼= PD and get that4APL∼=4DPM by SAS. In particular, we get ∠LAP∼=∠MDP,which forces ABCD to be isosceles.

7.6 No, α > β implies that BC is longer than AD.

7.7(a) Since AD||BC then ∠BAE ∼= ∠FCD. Then,4BEA∼=4DCF by SAS. It follows that ∠CDF = α .

(b)

7.8(a) (b)

Chapter 8

8.1 Assume that ABC, ACE, and CDE are semicircles, and that AC∼=CE. We want to show that the area ofthe shaded region is equal to the area of4ACE

D

A E

CB

98 14 Solutions

Since the vertices of4AEC are on the circle, then the height of4AEC is r and its base is 2r, where r isthe radius of the circle through A,E and C. It follows that the area of 4AEC is r·(2r)

2 = r2 and that (usingthe Pythagorean theorem with the altitude of4AEC) r2 + r2 = AC2. So, AC = r

√2.

The area of the whole figure is equal to the area of the triangle plus the area of the semicircles, we get

r2 +π

(r√

22

)2

= r2(

1+π

2

)Since the shaded area is obtained from subtracting the white semi-circle from the whole thing, we get

that the area of the shaded area isr2(

1+π

2

)− 1

2πr2 = r2

So, the area of the triangle equals the area of the shaded region.

8.2 Note that the area of the red region is the area of the circle minus the area of the white square (the blueparts are irrelevant at this point). So let r be the radius of the big circle. It follows that the diagonal of thesquare is 2r, which implies that the side of the square is a = r

√2. So, the red region has area

πr2− (r√

2)2 = r2(π−2)

The are of the blue region is found by looking at the bottom part, which is

The area of the rectangle is is r2 because it is half of the area of the square, the area of the semicircle is

12

π

(r√

22

)2

=πr2

4

because it is half of the area of a circle with radius r√

22 . Finally the area of the triangle in the figure is half

of the area of the rectangle. So, the area of the two top pieces (above the triangle) in blue is

πr2

4− r2

2=

r2

4(π−2)

Since the blue figure has 8 parts like the TWO we have found the area of then the blue area is

4r2

4(π−2) = r2(π−2)

8.3

8.4 Assuming that CXEY is a square, we want to show that the arc QP is congruent to the arc JT .First join the center with Q and J to get4QCJ

14 Solutions 99

P T

E

C

X Y

QJ

Note that 4CY J ∼= 4CXQ because both are right triangles, QC ∼= JC, and CX ∼= CY . This impliesthat QX ∼= JY . Since XE ∼= EY then QE ∼= JE. It follows that 4QJE is isosceles with base QJ, and thus∠QJP∼= ∠T QJ, which implies that arc QP is congruent to arc JT .

8.5 Since the sides of the triangle are tangent to the circle then two consecutive sides can be considered astangents coming out of the same point (the vertex). It follows that these two segments from the vertex to thecircle are congruent. Hence,

3

82

2

38and thus x = 11.

8.6 The picture for this is something like

2

x

8

8

Since a tangent is always perpendicular to the radius at the tangent point, then we can use the Pythagoreantheorem

82 + x2 = 102

which implies x = 6.

8.7 The square has perimeter 36 in and thus its side has length 6 in. Since the circle is circumscribedabout the square then the diameter of the circle is equal to the length a diagonal of the square. Using thePythagorean theorem with two sides of the square and the diagonal we get 62 +62 = d2, and thus d = 6

√2.

It follows that the circumference of the circle is 6√

2π in.

8.8

Chapter 9

9.1 Let the edges of the box have length x, 2x and 3x. Then the main diagonal in the box has length√x2 +(2x)2 +(3x)2 = 7

√2

100 14 Solutions

It follows that 14x2 = 98, and thus x =√

7. The volume of the box is√

7 · (2√

7) · (3√

7) = 42√

7

9.2 We cut the cylinder into two smaller cylinders, each one containing a full cone (half of the sand-clock).We know that the volume of a cone inscribed in this way in a cylinder is a third of the volume of thecylinder. It follows that the total volume of the cylinder is the sum of the volumes of the cylinders, and thusthe volume of the sand-clock is the sum of the volumes of the cones. Since each cone’s volume is a third ofthe smaller cylinders’, then the volume of the sand-clock is a third of the volume of the large cylinder.

9.3 We need to find the surface area of the can.The top and bottom have area π(22) = 4π cm2 each.The side of the cylinder has area 2π(4)(7) = 56π cm2.It follows that the can has total surface area 64π cm2.

9.4 Drop the altitude from the tip of the larger cone to the center of the base (this line also goes throughthe small cone’s center). Note that the slant(line), the altitude and the radii of the bases form the followingfigure.

R

15

4

8

r

where R is the radius of the large cone’s base and r the radius of the small cone’s base.Using the Pythagorean theorem we find that R = 9, and then using similarity of triangles we get that

r = 3 and that the slant of the smaller cone is 5.Since the larger cone has been obtained from the smaller one by zooming ×3 then the volume of the

larger cone is 27 times the volume of the smaller cone.

9.5 This is like a 3−D Pythagorean theorem. Take a face of the cube. We know the diagonal of that cube isy√

2 by using the Pythagorean theorem. Now the diagonal in the cube can be thought of as the hypothenuseof a triangle with legs a ‘vertical’ side of the cube and the diagonal of the square at the base of the cube.Thus, the Pythagorean theorem says.

x2 = y2 +(y√

2)2

which implies x = y√

3 in.Another way to do this is to use coordinates. We set one corner of the cube at (0,0,0) and the other must

be at (y,y,y). So, the distance between these points is

d =√(y−0)2 +(y−0)2 +(y−0)2 =

√3y2 = y

√3

9.6 Since the radius is a zoomed×3 then all lengths are multiplied by 3, and thus all areas will be multipliedby 32 and all volumes by 33 = 27.

9.7 The volume of the box is 3 ·8 ·9 = 216 in3. Since 63 = 216, then the side of the cube is 6 inches, andthus its side has area 36 in2.

9.8

14 Solutions 101

9.9

Chapter 10

10.1 We can see that the coordinates of the points are A = (−2,2) and B = (3,−1). It follows that the slopeof the line that joins them is

−1−23− (−2)

=− 35

The equation of the line through A and B can be found by using that A is a point of

y =− 35

x+b

We get b = 45 , and thus the line has equation

y =− 35

x+45

A line perpendicular to the line above must look like

y =53

x+b

Since (1,2) is on this line then b = 13 and thus

y =53

x+13

The intersection of these two lines is found by solving the system of linear equations

y =− 35

x+45

y =53

x+13

which we re-write5y =−3x+4 3y = 5x+1

This system has solution(

734

,2334

).

10.2

10.3

10.4 Set a vertex of the rectangle to be the origin of the plane, and two of its sides to lie on the x-axis andy-axis. It follows that the four vertices of the rectangle are (0,0),(a,0),(a,b), and (b,0).

The slopes of the lines that ‘contain’ the diagonals of the rectangle are:

m(0,0)−(a,b) =ba

m(a,0)−(0,b) =−ba

Since we are assuming the diagonal are perpendicular then the product of these slopes must be equal to−1. We get,

a2 = b2

and thus a =±b. But since both a and b are positive, then a = b and thus the rectangle is a square.

102 14 Solutions

10.5

10.6 With the same setting as the solution to problem 10.4. We first find the equations of the lines that‘contain’ the two diagonals. We already know the slopes, and also their y-intercepts, so it is easy to see thatthe lines are

y =ba

x y =− ba

x+b

These lines intersect at(

a2,

b2

), which is the midpoint of both of the diagonals.

10.7

10.8 (a) The slope of the line is

m =(a−1)− (a+1)(3a+5)−a

=−2

2a+5

Now using that A = (a,a+1) is on the line we get

a+1 =−2

2a+5(a)+b

which forces b =2a2 +9a+5

2a+5. And thus, the equation of the line is

y =−2

2a+5x+

2a2 +9a+52a+5

(b) Any line perpendicular to the line above must have slope2a+5

2. So, if the line goes through A =

(a,a+1) then

a+1 =2a+5

2(a)+b

which implies b =2−2a2−3a

2. And thus, the line asked for is

y =2a+5

2x+

2−2a2−3a2

(c) As mentioned above, this line must have slope2a+5

2. So, if the line goes through A = (3a+5,a−1)

thena−1 =

2a+52

(3a+5)+b

which implies b =−6a2−23a−27

2. And thus, the line asked for is

y =2a+5

2x+−6a2−23a−27

2

Chapter 11

11.1 We are given 4x2 +9y2 = 36. We divide by 36 to get

14 Solutions 103

x2

9+

y2

4= 1

It follows that the ellipse is centered at (0,0) and a = 3 and b = 2. A picture follows

Finally, since a2 = c2 +b2, then c =√

5, and thus the foci are at (±√

5,0).

11.2 This is the same as above, because we are just interchanging x and y. So, the ellipse is ‘vertical’centered at (0,0), and the foci are (0,±

√5).

11.3

11.4 It is clear that the center of the ellipse is (1,−1) and that a =√

5 and b =√

3. Since a2 = c2+b2, thenc =√

2. It follows that the foci are at (1±√

2,−1).

11.5 We find the center of the circle by finding the midpoint of the diameter given.(1+5

2,

1+42

)= (3,2.5)

Since the distance between (1,1) and (5,4) is√(5−1)2 +(4−1)2 =

√42 +32 = 5

then the radius of the circle is 2.5. It follows that the equation of the circle is

(x−3)2 +(y−2.5)2 = (2.5)2

11.6

11.7 x−h = 4p(y− k)2

11.8 We find the center of the circle by finding the midpoint of the diameter given.(3+6

2,

1+52

)= (4.5,3)

Since the distance between (3,1) and (6,5) is√(6−3)2 +(5−1)2 =

√32 +42 = 5

then the radius of the circle is 2.5. It follows that the equation of the circle is

(x−4.5)2 +(y−3)2 = (2.5)2

11.9

104 14 Solutions

Chapter 12

12.1 We notice that the translation maps (1,2) to (−2,6), thus the vector used for the translation is (−2,6)−(1,2) = (−2−1,6−2) = (−3,4). Note that we could have used any of the four corners of the rhombus toobtain this vector.

12.2 Since a reflection R across the x-axis maps a point (x,y) to (x,−y) then

R(−3,2) = (−3−2) R(−1,4) = R(−1,−4) R(−2,6) = (−2,−6)

It follows that A′ = (−3−2), B′ = R(−1,−4), and C′ = (−2,−6).

12.3

12.4

Chapter 13

13.1 Since the number of possible outcomes is 20, and there are 8 prime numbers less than 21 (2,3,5,7,11,13,17,and 19) then the probability is

P(X) =8

20=

25= 0.4

which is a 40% chance.

13.2 The total number of markers in my bag is 43, the total number of red markers is 12, thus

P(X) =1243∼ .28

which is a 28% chance.

13.3 The area of the square is 100, and thus this will be considered to be the ‘total number of outcomes’.The diameter of the circle equals the length of the side of the square, and thus its area is 52π = 25π . Hence,the probability asked for is

P(X) =25π

100=

π

4∼ 0.78

which is about a 78% chance.

13.4 Since months have not all the same number of days then in order to compute this we need to look atdays, not at months.

In this problem, the possible outcomes are the days of the year (no leap year), so there are 365 of them.The favorable outcomes are the days of the months ending in “ary”. These months are January (31 days)and February (28 days). So, there are 59 favorable outcomes. It follows that

P(X) =59

365∼ 0.16

which is approximately 16%.

13.5

14 Solutions 105

Chapter 14

14.1

14.2 If the data is distributed in a normal curve (bell-shaped) then the highest point of the curve is exactlyat the middle of the data values, that means that it is exactly where the median is. Moreover, since that isthe highest point then the frequency of the median is as large as possible, and thus the median is also themode. Finally, since the data is symmetrically distributed then whenever there is a value that is, let us say, 1unit less than the median, then there is another that balances it out because it must be 1 unit larger than themedian. This means that the mean is also equal to the median.

14.3

Index

χ2-test, 97, 98z-score, 96, 97

Angle Bisector, 9, 11, 37Arc, 56Arithmetic Mean, 90ASA, 36

Calculator, 100Central Angle, 54, 56Chord, 53, 56Circle, 53, 69Circle Section, 56Combinations, 86Common Notions, 13Congruency, 21, 35, 36Conic Sections, 69, 74Constructions, 4, 13Coordinate Geometry, 67Correlation, 98

Dihedral Angle, 63Dilation, 78Direct Proofs, 2Distance Formula, 67

Ellipse, 70, 71Equation Of The Line, 67Euclid, 13, 14Euclid’s Fifth Postulate, 14, 15, 19Euclidean Geometry, 7

Frequency, 89Fundamental Principle of Counting, 85

Girard’s Theorem, 107

Hyperbola, 72, 73

Inscribed Angle, 54Isometry, 77, 78Isosceles Trapezoid, 50

Kite, 47

Linear Regression, 99, 100

Median, 91Midpoint, 8, 11Midpoint Formula, 67Mode, 90

Neutral Geometry, 15Non-Euclidean Geometry, 14, 15, 19Normal Distribution, 90, 93Normal Vector, 63Null Hypothesis, 94

Parabola, 73Parallel Lines, 63Parallelogram, 47Percentile, 93Permutations, 86Perpendicular Bisector, 9, 11, 37Polygon, 25Polygons, 25, 26Postulates, 4, 14Prism, 63Probability, 81, 82, 87Proof by Contradiction, 3Proofs, 1

Quadrilaterals, 47

Radians, 8, 11, 54Rectangle, 47Reflection, 77Regular Polygons, 27Rhombus, 47Rotation, 77

SAS, 35Segment, 8Similarity, 21, 22, 38, 39, 42Skew Lines, 63Slope, 67SSS, 36Standard Deviation, 92Statistical Tables, 102, 103Statistics, 89Surface Area, 64

Thales’ theorem, 42Translation, 77, 79

107

108 Index

Trapezoid, 48Triangles, 31–36, 38, 39, 42Trigonometry, 39, 41, 42

Variance, 92Vertical Angle Theorem, 2, 9, 17Volume, 64